Titrations Questions in English

(B) Let the mass of $Na_2CO_3$ be $x \ g$ and $NaOH$ be $y \ g$.
Using phenolphthalein indicator,$NaOH$ is completely neutralized and $Na_2CO_3$ is converted to $NaHCO_3$ (half-neutralization).
$Meq. \ of \ NaOH + Meq. \ of \ Na_2CO_3 (half) = 300 \times 0.1 = 30$.
$\frac{y}{40} \times 1000 + \frac{x}{106} \times 1000 = 30$ ... $(1)$
When methyl orange is added,the remaining $NaHCO_3$ is neutralized.
$Meq. \ of \ Na_2CO_3 (remaining \ half) = 25 \times 0.2 = 5$.
$\frac{x}{106} \times 1000 = 5$ ... $(2)$
From $(2)$,$\frac{x}{106} \times 1000 = 5$.
Substitute this into $(1)$:
$\frac{y}{40} \times 1000 + 5 = 30$
$\frac{y}{40} \times 1000 = 25$
$y = \frac{25 \times 40}{1000} = 1 \ g$.

(A) The reaction between ferrous ammonium sulphate (Mohr's salt) and $KMnO_4$ in acidic medium is:
$MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O$.
From the stoichiometry,$1 \, \text{mole of } KMnO_4$ reacts with $5 \, \text{moles of } Fe^{2+}$.
Number of moles of $Fe^{2+}$ in $20 \, mL$ of solution:
Total moles in $100 \, mL = \frac{3.92}{392} = 0.01 \, \text{mol}$.
Moles in $20 \, mL = 0.01 \times \frac{20}{100} = 0.002 \, \text{mol}$.
Since $5 \, \text{moles of } Fe^{2+}$ react with $1 \, \text{mole of } KMnO_4$,$0.002 \, \text{mol of } Fe^{2+}$ react with $\frac{0.002}{5} = 0.0004 \, \text{mol of } KMnO_4$.
This amount is present in $18 \, mL$ of $KMnO_4$ solution.
Concentration of $KMnO_4$ in $g/L = \frac{0.0004 \, \text{mol}}{18 \, mL} \times 1000 \, mL \times 158 \, g/mol = 3.51 \, g/L$.
Given the options provided,the closest value is $3.476 \, g$.

(B) An organic dye,such as eosin,used to detect the endpoint of a precipitation titration through the process of adsorption is known as an $Adsorption$ $indicator$.
This technique is commonly employed in $Argentometry$,which is a type of titration involving the silver$(I)$ ion.
Typically,it is used to determine the amount of chloride present in a sample by titrating it against a silver nitrate solution of known concentration.
Chloride ions react with silver$(I)$ ions to form insoluble silver chloride:
$Ag^+_{(aq)} + Cl^-_{(aq)} \rightarrow AgCl_{(s)} \ (K = 5.88 \times 10^9)$
Fluorescein and eosin are common examples of adsorption indicators used in this reaction:
$NaCl_{(aq)} + AgNO_{3(aq)} \rightarrow AgCl_{(s)} + NaNO_{3(aq)}$

(A) In an iodometric titration,a starch solution is used as an indicator because it forms a blue-black complex with $I_2$.
As the titration proceeds,the $I_2$ is consumed by the sodium thiosulphate $(Na_2S_2O_3)$.
The disappearance of the blue colour marks the endpoint of the titration.

(B) The titration of $Na_2CO_3$ (a weak base) with $H_2SO_4$ (a strong acid) involves the formation of $NaHCO_3$ and then $H_2CO_3$.
Methyl orange is the suitable indicator for this titration because it changes color in the $pH$ range of $3.1$ to $4.4$,which corresponds to the acidic end point of the reaction.

(B) Strong acids are preferred as standard solutions because they react completely with both strong and weak bases,ensuring a sharp change in $pH$ near the equivalence point,which makes the detection of the endpoint easier. Thus,they can be used to titrate both strong and weak bases.

(A) For the neutralization of an acid and a base,the total milliequivalents $(N \times V)$ must be equal.
Total milliequivalents of $HCl = 0.1 \ N \times 100 \ mL = 10 \ meq$.
Milliequivalents of $NaOH$ added = $0.2 \ N \times 30 \ mL = 6 \ meq$.
Remaining milliequivalents of $HCl$ to be neutralized = $10 \ meq - 6 \ meq = 4 \ meq$.
Let the volume of $KOH$ required be $V \ mL$. Since the normality of $KOH$ is $0.25 \ N$,we have:
$0.25 \ N \times V \ mL = 4 \ meq$.
$V = \frac{4}{0.25} = 16 \ mL$.

(B) According to the law of equivalence,for neutralization reactions,the number of equivalents of acid equals the number of equivalents of base.
$N_1V_1 = N_2V_2$
Here,$N_1 = \frac{1}{10} \ N$,$V_1 = 8 \ mL$,and $V_2 = 20 \ mL$.
Substituting the values:
$\frac{1}{10} \times 8 = N_2 \times 20$
$0.8 = N_2 \times 20$
$N_2 = \frac{0.8}{20} = 0.04 \ N$
Therefore,the normality of the $Na_2CO_3$ solution is $0.04 \ N$.

(D) In the titration of a strong acid and a weak base,the indicator used is Methyl orange.
Methyl orange is an indicator that is used to determine the equivalence point of an acid-base titration where the $pH$ at the equivalence point is in the acidic range (typically $3.1$ to $4.4$).
In an acidic environment,it turns red,and in a basic environment (or $pH$ of $4.4$ and higher),it turns yellow.
It is prepared by the treatment of helianthin with sodium hydroxide; helianthin is obtained by coupling diazotized sulphanilic acid with $N,N$-dimethylaniline.

(B) An organic dye,Eosin,used to detect the endpoint of precipitation titration by adsorption,is known as an adsorption indicator.
This process is also referred to as argentometric titration.

(A) The neutralization reaction is: $HCl + NaOH \rightarrow NaCl + H_2O$.
According to the stoichiometry,$1 \ mole$ of $HCl$ reacts with $1 \ mole$ of $NaOH$.
Using the titration formula $M_1V_1 = M_2V_2$:
Let $M_1$ be the molarity of $HCl$ and $V_1 = 20 \ mL$.
Let $M_2 = 0.01 \ M$ be the molarity of $NaOH$ and $V_2 = 19.85 \ mL$.
$M_1 \times 20 = 0.01 \times 19.85$
$M_1 = \frac{0.1985}{20} = 0.009925 \ M$.
Rounding to the appropriate significant figures,the molarity is $0.0099 \ M$.

(A) The balanced chemical equation for the reaction is: $Ba(OH)_2 + 2HCl \rightarrow BaCl_2 + 2H_2O$.
According to the stoichiometry,$1 \ mole$ of $Ba(OH)_2$ reacts with $2 \ moles$ of $HCl$.
Using the molarity equation: $\frac{M_1 V_1}{n_1} = \frac{M_2 V_2}{n_2}$,where $M_1$ and $V_1$ are for $Ba(OH)_2$ and $M_2$ and $V_2$ are for $HCl$.
Given: $V_1 = 25 \ mL$,$M_2 = 0.1 \ M$,$V_2 = 35 \ mL$,$n_1 = 1$,$n_2 = 2$.
$\frac{M_1 \times 25}{1} = \frac{0.1 \times 35}{2}$.
$M_1 = \frac{0.1 \times 35}{25 \times 2} = \frac{3.5}{50} = 0.07 \ M$.

(A) The equivalent weight of ferrous ammonium sulfate is $392 \, g/eq$.
The normality of the salt solution is $N_1 = \frac{3.92}{392} \times \frac{1000}{100} = 0.1 \, N$.
Using the titration formula $N_1V_1 = N_2V_2$ for $20 \, mL$ of salt solution and $18 \, mL$ of $KMnO_4$:
$0.1 \times 20 = N_2 \times 18$.
Normality of $KMnO_4$ solution $(N_2)$ $= \frac{2}{18} = \frac{1}{9} \, N$.
The equivalent weight of $KMnO_4$ in acidic medium is $31.6 \, g/eq$.
Strength of $KMnO_4$ in $g/L = \text{Normality} \times \text{Equivalent weight} = \frac{1}{9} \times 31.6 \approx 3.51 \, g/L$.

(A) Phenolphthalein is a weak organic acid. In the titration of $Na_2CO_3$ with $HCl$,phenolphthalein changes color at the first equivalence point where $Na_2CO_3$ is converted to $NaHCO_3$. At this point,the solution becomes colorless.

(C) Let the normality of the diluted $HCl$ solution be $N_1$ and its volume be $V_1 = 20 \, mL$.
For $NaOH$,$N_2 = 0.1 \, N$ and $V_2 = 25 \, mL$.
Using the titration formula $N_1 V_1 = N_2 V_2$:
$N_1 \times 20 = 0.1 \times 25$
$N_1 = \frac{0.1 \times 25}{20} = 0.125 \, N$.
This $N_1$ is the normality of the diluted solution $(1 \, L)$.
To find the normality of the original $10 \, mL$ concentrated $HCl$,we use the dilution equation $N_{conc} V_{conc} = N_{dil} V_{dil}$:
$N_{conc} \times 10 \, mL = 0.125 \, N \times 1000 \, mL$
$N_{conc} = \frac{0.125 \times 1000}{10} = 12.5 \, N$.

(C) Methyl orange is a weak base indicator with a $pH$ range of approximately $3.1$ to $4.4$.
It is best suited for titrations involving a strong acid and a weak base,or a strong acid and a strong base where the equivalence point falls within its $pH$ range.
Among the given options,$HCl$ is a strong acid and $NaOH$ is a strong base,resulting in a neutral salt where the equivalence point is at $pH = 7$. However,methyl orange is commonly used for strong acid-strong base titrations in educational contexts.
$CH_3COOH + NH_4OH$ involves a weak acid and a weak base,which is not suitable for simple indicators.
Therefore,the titration of a strong acid $(HCl)$ with a strong base $(NaOH)$ is the most appropriate choice among the options provided for the use of methyl orange.

(D) Phenolphthalein is a weak organic acid indicator with a pH range of $8.2$ to $10.0$. It is suitable for titrations involving strong bases and strong acids,or weak acids and strong bases.
$HCl$ is a strong acid.
$NaOH$,$RbOH$,and $KOH$ are all strong bases.
$NH_4OH$ is a weak base.
Titration of a strong acid $(HCl)$ with a weak base $(NH_4OH)$ results in an equivalence point in the acidic range (pH $< 7$). Since phenolphthalein changes color in the basic range (pH $8.2-10.0$),it is unsuitable for this titration.

(A) In iodometric titration,an oxidizing agent reacts with excess iodide ions $(I^-)$ in an acidic or neutral medium to liberate iodine $(I_2)$.
The liberated iodine $(I_2)$ is then titrated against a standard solution of sodium thiosulfate $(Na_2S_2O_3)$.
The reaction $Cr_2O_7^{2-} + 14H^+ + 6I^- \to 2Cr^{3+} + 3I_2 + 7H_2O$ represents the liberation of $I_2$,followed by the titration reaction $I_2 + 2S_2O_3^{2-} \to S_4O_6^{2-} + 2I^-$.
Therefore,option $A$ correctly describes the iodometric titration process.

(B) The titration of oxalic acid with $KMnO_4$ is performed in the presence of $H_2SO_4$ because $H_2SO_4$ provides an acidic medium without interfering with the redox reaction.
If $HCl$ is used,$KMnO_4$ acts as a strong oxidizing agent and oxidizes $HCl$ to $Cl_2$ gas:
$2KMnO_4 + 16HCl \to 2KCl + 2MnCl_2 + 8H_2O + 5Cl_2$.
Due to this side reaction,the consumption of $KMnO_4$ increases,leading to an inaccurate determination of the oxalic acid concentration.

(B) The disodium salt of $Ethylenediaminetetraacetic$ $acid$ $(EDTA)$ is used for the complexometric titration of metal ions like $Ca^{2+}$ and $Mg^{2+}$ to determine water hardness.
In this titration,an indicator $(D^{3-})$ forms a red metal-indicator complex with $M^{2+}$ ions.
$i.$ $M^{2+} + D^{3-} \rightarrow [MD]^{-}$ (Red complex)
$ii.$ $[MD]^{-} + EDTA^{4-} \rightarrow [M(EDTA)]^{2-} + D^{3-}$ (Blue free indicator)
The $EDTA$ forms a more stable complex with the metal ions,displacing the indicator and causing a color change from red to blue at the endpoint.

(A) The titration of oxalic acid with $KMnO_4$ in the presence of $HCl$ gives an unsatisfactory result because $KMnO_4$ is a strong oxidizing agent and it oxidizes $HCl$ to $Cl_2$ gas along with the oxidation of oxalic acid.
This leads to an error in the titration as the volume of $KMnO_4$ consumed is higher than that required for the oxidation of oxalic acid alone.

(D) $1$. $Ce^{+4}$ is a strong oxidising agent because it easily reduces to $Ce^{+3}$. It does not act as a reducing agent.
$2$. Pure $PH_3$ is prepared by heating $PH_4I$ with a strong base like $KOH$ or $NaOH$,not $CH_3COOH$.
$3$. Urea,sulphamic acid,and thiourea are used to destroy nitrite ions $(NO_2^-)$,not nitrate ions $(NO_3^-)$.
$4$. Boric acid is a very weak acid $(K_a \approx 5.8 \times 10^{-10})$,and its titration with $NaOH$ is not satisfactory because the end point is not sharp. It requires the addition of polyols like glycerol to form a strong complex acid.

(A) The titration involves two steps:
$1$. $HCl$ is a strong acid and reacts first with $NaOH$: $H^+ + Cl^- + Na^+ + OH^- \rightarrow H_2O + Na^+ + Cl^-$. The highly mobile $H^+$ ions are replaced by $Na^+$ ions,leading to a decrease in conductance.
$2$. After $HCl$ is neutralized,$HCN$ (a weak acid) reacts with $NaOH$: $HCN + Na^+ + OH^- \rightarrow CN^- + Na^+ + H_2O$. The concentration of ions increases as $HCN$ is converted to $Na^+ + CN^-$,leading to a slight increase in conductance.
$3$. After both acids are neutralized,adding excess $NaOH$ adds $Na^+$ and $OH^-$ ions,causing a sharp increase in conductance.
Therefore,the graph shows a decrease,then a slight increase,followed by a sharp increase. This matches the pattern in Graph $A$.

(NONE) The reaction between $Ca(HCO_3)_2$ and $H_2SO_4$ is: $Ca(HCO_3)_2 + H_2SO_4 \rightarrow CaSO_4 + 2H_2O + 2CO_2$.
At the equivalence point,the milliequivalents of $Ca(HCO_3)_2$ equal the milliequivalents of $H_2SO_4$.
$n_{eq}(Ca(HCO_3)_2) = n_{eq}(H_2SO_4) = M \times n_{factor} \times V(L) = \frac{1}{20} \times 2 \times 0.040 = 0.004 \ eq$.
Since $1 \ mol$ of $Ca(HCO_3)_2$ is equivalent to $1 \ mol$ of $CaCO_3$ $(100 \ g/mol)$,the mass of $CaCO_3$ is $0.004 \ mol \times 100 \ g/mol = 0.4 \ g$.
Hardness in $ppm = \frac{\text{mass of } CaCO_3 \text{ in } g}{\text{volume of water in } mL} \times 10^6 = \frac{0.4}{500} \times 10^6 = 800 \ ppm$.

(B) Step $1$: Calculate the initial millimoles of $Ba(OH)_2$.
$Molarity = 0.1 \ M$,$Volume = 300 \ mL$.
$Millimoles \ of \ Ba(OH)_2 = 0.1 \times 300 = 30 \ mmol$.
$Milliequivalents \ of \ Ba(OH)_2 = 30 \times 2 = 60 \ meq$.
Step $2$: Calculate the milliequivalents of $HCl$ added.
$Molarity = 0.2 \ M$,$Volume = 100 \ mL$.
$Milliequivalents \ of \ HCl = 0.2 \times 100 \times 1 = 20 \ meq$.
Step $3$: Calculate remaining $Ba(OH)_2$ after reaction with $HCl$.
$Remaining \ meq \ of \ Ba(OH)_2 = 60 - 20 = 40 \ meq$.
Step $4$: Titrate remaining $Ba(OH)_2$ with $H_2SO_4$.
$Normality \ of \ H_2SO_4 = 0.5 \ M \times 2 = 1 \ N$.
$Equivalence \ point: meq \ of \ Ba(OH)_2 = meq \ of \ H_2SO_4$.
$40 = 1 \times V$.
$V = 40 \ mL$.

(A) In the titration of oxalic acid against $KMnO_4$,the reaction occurs in an acidic medium (using $H_2SO_4$).
The chemical reaction is:
$2KMnO_4 + 3H_2SO_4 + 5(COOH)_2 \rightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 10CO_2 \uparrow$
Potassium permanganate $(KMnO_4)$ acts as a strong oxidizing agent and is deep purple in color.
As the titration proceeds,the $KMnO_4$ added is consumed by the oxalic acid. Once all the oxalic acid has reacted,the very next drop of $KMnO_4$ added remains unreacted,imparting a permanent light pink color to the solution.
Therefore,$KMnO_4$ itself acts as a self-indicator.

(C) primary standard is a reagent that is pure,stable,and has a high molar mass.
Oxalic acid $(H_2C_2O_4 \cdot 2H_2O)$ is a solid,stable compound that meets these criteria and is commonly used to standardize $NaOH$ solutions via titration.

(D) The molar mass of oxalic acid dihydrate $(H_2C_2O_4 \cdot 2H_2O)$ is $126 \ g/mol$.
Since the basicity of oxalic acid is $2$,its equivalent mass is $126 / 2 = 63 \ g/eq$.
Normality $(N)$ of the oxalic acid solution is calculated as:
$N = \frac{\text{mass}}{\text{equivalent mass} \times \text{volume in L}} = \frac{6.3}{63 \times 0.250} = 0.4 \ N$.
Using the law of equivalence $(N_1 V_1 = N_2 V_2)$ for neutralization:
$0.4 \ N \times 10 \ mL = 0.1 \ N \times V_2$.
$V_2 = \frac{0.4 \times 10}{0.1} = 40 \ mL$.

(A) Step $1$: Determine the normality of $HCl$ using the reaction with $Na_2CO_3$.
$Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$.
Using the law of equivalence: $N_1V_1 = N_2V_2$.
For $Na_2CO_3$,$N = M \times \text{valency factor} = 0.1 \times 2 = 0.2 \ N$.
$N_{HCl} \times 25 \ mL = 0.2 \ N \times 30 \ mL$.
$N_{HCl} = \frac{6}{25} = 0.24 \ N$.
Step $2$: Determine the volume of $HCl$ required for $NaOH$.
$HCl + NaOH \rightarrow NaCl + H_2O$.
Using $N_1V_1 = N_2V_2$ for $HCl$ and $NaOH$.
$0.24 \ N \times V_{HCl} = 0.2 \ M \times 1 \times 30 \ mL$.
$0.24 \times V_{HCl} = 6$.
$V_{HCl} = \frac{6}{0.24} = 25 \ mL$.

(A) In this titration,a strong acid $(HCl)$ is added to a strong base $(NaOH)$.
Initially,the solution contains $NaOH$,so the $pH$ is high (basic).
As $HCl$ is added,the $pH$ decreases gradually.
Near the equivalence point,there is a sharp drop in $pH$ because the concentration of $OH^-$ ions decreases rapidly.
After the equivalence point,the solution becomes acidic due to excess $HCl$,and the $pH$ levels off at a low value.
Graph $(a)$ correctly represents this sigmoidal decrease in $pH$ as the volume of the acid titrant increases.

(A) In the titration of oxalic acid with $KMnO_4$,the reaction is a redox titration.
$KMnO_4$ acts as both the oxidizing agent and the self-indicator.
Initially,the solution is colorless due to oxalic acid.
As $KMnO_4$ is added,it reacts with oxalic acid to form colorless $Mn^{2+}$ ions.
At the equivalence point,the very first drop of excess $KMnO_4$ imparts a permanent light pink color to the solution,signaling the end point.
Therefore,$KMnO_4$ acts as a self-indicator.

(B) The total milliequivalents of $HCl$ must equal the sum of the milliequivalents of $NaOH$ and $KOH$ used for titration.
$N_1V_1 = N_2V_2 + N_3V_3$
Substituting the given values:
$0.1 \times 110 = (0.2 \times 30) + (0.25 \times V_{KOH})$
$11 = 6 + 0.25 \times V_{KOH}$
$11 - 6 = 0.25 \times V_{KOH}$
$5 = 0.25 \times V_{KOH}$
$V_{KOH} = \frac{5}{0.25} = 20 \, mL$

(A) The neutralization reaction is: $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$.
Using the titration formula $n_1 M_1 V_1 = n_2 M_2 V_2$,where $n$ is the acidity/basicity factor.
For $H_2SO_4$,$n_1 = 2$ and for $NaOH$,$n_2 = 1$.
Substituting the values: $2 \times M_1 \times 25 \ mL = 1 \times 0.164 \ M \times 32.63 \ mL$.
$M_1 = \frac{0.164 \times 32.63}{2 \times 25} = \frac{5.35132}{50} = 0.107 \ M$.

(C) The hardness of water is determined by complexometric titration using the disodium salt of $EDTA$ (Ethylenediaminetetraacetic acid).
In this process,$EDTA$ forms stable complexes with metal ions like $Ca^{2+}$ and $Mg^{2+}$ present in the water,which are responsible for water hardness.
Eriochrome Black $T$ $(EBT)$ is typically used as an indicator in this titration.

(B) The molar mass of oxalic acid dihydrate $(H_2C_2O_4 \cdot 2H_2O)$ is $126 \ g/mol$.
Since it is a dibasic acid,its equivalent mass is $\frac{126}{2} = 63 \ g/eq$.
Normality of the oxalic acid solution = $\frac{\text{mass}}{\text{equivalent mass} \times \text{volume in L}} = \frac{6.3}{63 \times 0.250} = 0.4 \ N$.
Using the titration formula $N_1 V_1 = N_2 V_2$ for $10 \ mL$ of the solution:
$0.4 \ N \times 10 \ mL = 0.1 \ N \times V_2$.
$V_2 = \frac{4}{0.1} = 40 \ mL$.

(B) The strength of an aqueous solution of $I_2$ is determined by iodometric titration.
In this process,$I_2$ reacts with sodium thiosulphate $(Na_2S_2O_3)$ to form sodium tetrathionate $(Na_2S_4O_6)$ and sodium iodide $(NaI)$.
The balanced chemical equation is:
$I_2 + 2Na_2S_2O_3 \to Na_2S_4O_6 + 2NaI$

(D) The strength of a base like $NaOH$ is determined by titration against a primary standard acid.
Oxalic acid $(H_{2}C_{2}O_{4} \cdot 2H_{2}O)$ is a primary standard,meaning it can be weighed accurately to prepare a solution of known concentration.
Strong acids like $H_{2}SO_{4}$ are not primary standards because they are hygroscopic and their concentration changes over time.
In a titration,the solution with the known concentration (standard solution) is typically placed in the burette,but for $NaOH$ titration,it is standard practice to place the $NaOH$ solution in the burette and the primary standard oxalic acid solution in the conical flask.
Therefore,the correct setup is $NaOH$ in the burette and aqueous oxalic acid in the conical flask.

(A) No,titrations with $KMnO_4$ cannot be performed in the presence of $HCl$.
$HCl$ is a reducing agent and it reacts with $KMnO_4$ to produce $Cl_2$ gas.
The reaction is: $2KMnO_4 + 16HCl \rightarrow 2KCl + 2MnCl_2 + 8H_2O + 5Cl_2$.
This reaction interferes with the titration process by consuming the titrant and producing a gas,leading to inaccurate results.

(N/A) The amount of excess alkali in soap can be determined by performing an acid-base titration using a standard acid solution with a suitable indicator.
The source of excess alkali is typically the unreacted $NaOH$ or $KOH$ that remains in the soap mixture after the saponification process (hydrolysis of fat/oil) is complete.

(N/A) The titration of $KMnO_{4}$ and oxalate is not performed at room temperature; instead,a temperature range of $50^{\circ}C - 70^{\circ}C$ is maintained. At room temperature,the rate of reaction is extremely slow,and the reaction is autocatalyzed by the $Mn^{2+}$ ions produced during the process.

(B) The reaction between $KMnO_4$ and oxalic acid is very slow at room temperature.
Heating the solution provides the necessary activation energy and increases the frequency of effective collisions,thereby increasing the rate of the reaction.

(D) In an acid-base titration between $HCl$ and $NaOH$,the following apparatus and reagents are typically used:
$1$. $A$ $Burette$ to hold the titrant $(NaOH)$.
$2$. $A$ $Pipette$ to measure a precise volume of the analyte $(HCl)$.
$3$. $A$ $Clamp$ stand to hold the burette.
$4$. $Phenolphthalein$ as an acid-base indicator.
$5$. $Distilled water$ for rinsing and preparing solutions.
$6$. $A$ $Porcelain tile$ is often placed under the conical flask to observe the color change clearly.
$7$. $A$ $Bunsen burner$ and a $measuring cylinder$ are not required for this titration as it is performed at room temperature and requires precise volumetric glassware rather than a measuring cylinder.

(C) The most consistent and precise volume of $HCl$ obtained from the readings is $5.0 \, mL$.
At the equivalence point,the number of milliequivalents $(meq)$ of $Na_{2}CO_{3}$ equals the number of milliequivalents of $HCl$.
Let the molarity of $Na_{2}CO_{3}$ solution be $M$.
The reaction is: $Na_{2}CO_{3} + 2HCl \rightarrow 2NaCl + H_{2}O + CO_{2}$.
The n-factor for $Na_{2}CO_{3}$ is $2$ and for $HCl$ is $1$.
Using the formula: $M_{1} \times V_{1} \times n_{1} = M_{2} \times V_{2} \times n_{2}$
$M \times 10 \, mL \times 2 = 0.2 \, M \times 5.0 \, mL \times 1$
$20 \times M = 1.0$
$M = 0.05 \, mol/L$
To convert to $mM$ (millimolar),we multiply by $1000$:
$0.05 \times 1000 = 50 \, mM$.

(B) Let $x$ be the milliequivalents of $NaOH$ and $y$ be the milliequivalents of $Na_{2}CO_{3}$.
At the first end point (phenolphthalein indicator),$NaOH$ is neutralized and $Na_{2}CO_{3}$ is converted to $NaHCO_{3}$:
$x + y = \frac{1}{10} \times 17.5 = 1.75 \text{ meq} \dots (1)$
At the second end point (methyl orange indicator),$NaHCO_{3}$ is neutralized:
$y = \frac{1}{10} \times 1.5 = 0.15 \text{ meq} \dots (2)$
Mass of $Na_{2}CO_{3} = \text{milliequivalents} \times \text{equivalent weight} \times 10^{-3}$
$= 0.15 \times 53 \times 10^{-3} = 0.00795 \ g$
Weight percentage of $Na_{2}CO_{3} = \frac{0.00795}{0.4} \times 100 = 1.9875 \%$
Rounding off to the nearest integer,we get $2 \%$.

(A) The concordant volume of $NaOH$ is the average of the consistent readings $(iii, iv, v)$,which is $4.4\, mL$.
Using the principle of equivalence: $\text{equivalents of } NaOH = \text{equivalents of } H_2C_2O_4$.
$M_{NaOH} \times V_{NaOH} \times n_{factor} = M_{acid} \times V_{acid} \times n_{factor}$.
For $NaOH$,$n_{factor} = 1$. For oxalic acid $(H_2C_2O_4)$,$n_{factor} = 2$.
$M \times 4.4 \times 1 = 1.25 \times 10.0 \times 2$.
$M = \frac{25}{4.4} \approx 5.68\, M$.
Rounding to the nearest integer,we get $6\, M$.

(B) The balanced redox reaction is: $6Fe^{2+} + Cr_{2}O_{7}^{2-} + 14H^{+} \rightarrow 6Fe^{3+} + 2Cr^{3+} + 7H_{2}O$.
At the equivalence point,the milli-equivalents of $Fe^{2+}$ equal the milli-equivalents of $K_{2}Cr_{2}O_{7}$.
For $Fe^{2+}$,the n-factor is $1$ $(Fe^{2+} \rightarrow Fe^{3+} + e^{-})$.
For $K_{2}Cr_{2}O_{7}$,the n-factor is $6$ $(Cr_{2}O_{7}^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_{2}O)$.
Let $M$ be the molarity of $Fe^{2+}$ solution.
$M \times 10 \times 1 = 0.02 \times 15 \times 6$.
$10M = 1.8$.
$M = 0.18 \ M = 18 \times 10^{-2} \ M$.
Thus,the value of $x$ is $18$.

(D) The balanced chemical equation for the reaction is:
$2KMnO_4 + 5H_2C_2O_4 + 3H_2SO_4 \rightarrow K_2SO_4 + 2MnSO_4 + 10CO_2 + 8H_2O$
At equivalence point,the number of equivalents of $KMnO_4$ equals the number of equivalents of oxalic acid dihydrate $(H_2C_2O_4 \cdot 2H_2O)$:
$n_{eq} (KMnO_4) = n_{eq} (H_2C_2O_4 \cdot 2H_2O)$
Using the formula $n_{eq} = M \times V(L) \times n$-factor:
$n$-factor for $KMnO_4 = 5$ (reduction of $Mn^{+7}$ to $Mn^{+2}$)
$n$-factor for $H_2C_2O_4 \cdot 2H_2O = 2$ (oxidation of $C^{+3}$ to $C^{+4}$)
$0.05 \times 10 \times 5 = M_{oxalic} \times 10 \times 2$
$2.5 = 20 \times M_{oxalic}$
$M_{oxalic} = 0.125 \, M$
Molar mass of $H_2C_2O_4 \cdot 2H_2O = 126 \, g/mol$
Strength in $g/L = Molarity \times Molar \, mass = 0.125 \times 126 = 15.75 \, g/L$
Converting to the required form: $15.75 \, g/L = 1575 \times 10^{-2} \, g/L$.

(B) From the stoichiometry of the reactions:
$2 Cu^{2+} \equiv I_{2} \equiv 2 S_{2}O_{3}^{2-}$.
Therefore,$n_{\text{eq.}}$ of $Cu^{2+} = n_{\text{eq.}}$ of $S_{2}O_{3}^{2-}$.
$n_{\text{eq.}}$ of $S_{2}O_{3}^{2-} = M \times V \times n_{\text{factor}} = 0.02 \times 20 \times 1 = 0.4 \, \text{mmol}$.
Since $n_{\text{eq.}}$ of $Cu^{2+} = 0.4 \, \text{mmol}$ and $n_{\text{factor}}$ for $Cu^{2+}$ is $1$ (change in oxidation state from $+2$ to $+1$ in $Cu_{2}I_{2}$),we have $n_{\text{mol}}$ of $Cu^{2+} = 0.4 \, \text{mmol}$.
$[Cu^{2+}] = \frac{n_{\text{mol}}}{V_{\text{solution}}} = \frac{0.4 \, \text{mmol}}{10 \, mL} = 0.04 \, M = 4 \times 10^{-2} \, M$.
Thus,the value is $4$.

(A) At the equivalence point,the number of equivalents of $FeCl_{2}$ equals the number of equivalents of $K_{2}Cr_{2}O_{7}$.
$N_{1}V_{1} = N_{2}V_{2}$
First,calculate the normality of the $K_{2}Cr_{2}O_{7}$ solution:
$n$-factor of $K_{2}Cr_{2}O_{7} = 6$ (since $Cr_{2}O_{7}^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_{2}O$).
Equivalent mass of $K_{2}Cr_{2}O_{7} = \frac{294}{6} = 49 \, g \, eq^{-1}$.
Normality $(N) = \frac{\text{mass}}{\text{eq. mass} \times \text{volume in } L} = \frac{1.225}{49 \times 0.250} = 0.1 \, N$.
Now,using the titration formula $N_{FeCl_{2}} \times V_{FeCl_{2}} = N_{K_{2}Cr_{2}O_{7}} \times V_{K_{2}Cr_{2}O_{7}}$:
$N_{FeCl_{2}} \times 10 \, mL = 0.1 \, N \times 25 \, mL$.
$N_{FeCl_{2}} = \frac{0.1 \times 25}{10} = 0.25 \, N$.