Acid and base indicators Questions in English

(C) $pH$ indicators are organic substances that are either weak acids or weak bases.
They exhibit different colors in their ionized and unionized forms.
The color change occurs due to the change in the concentration of $H^+$ ions in the solution,which shifts the equilibrium of the indicator.

(A) In the titration of a strong acid and a weak base,the equivalence point lies in the acidic range (typically $pH < 7$).
Methyl orange is a suitable indicator for this titration because its $pH$ range $(3.1 - 4.4)$ falls within the acidic region where the equivalence point occurs.
Therefore,the correct option is $(A)$.

(B) Phenolphthalein is an acid-base indicator that changes color in the $pH$ range of $8.2$ to $10.0$.
It is suitable for titrations involving strong bases and weak acids or strong bases and strong acids.
Option $B$ involves a redox titration between oxalic acid $(H_2C_2O_4)$ and potassium permanganate $(KMnO_4)$,where $KMnO_4$ acts as a self-indicator.
Therefore,phenolphthalein is not used as an indicator for this reaction.

(B) An indicator $(HIn)$ is a weak acid that dissociates in solution as follows:
$HIn \rightleftharpoons H^{+} + In^{-}$
The equilibrium constant expression is:
$K_{In} = \frac{[H^{+}][In^{-}]}{[HIn]}$
Taking the negative logarithm on both sides:
$-\log K_{In} = -\log [H^{+}] - \log \frac{[In^{-}]}{[HIn]}$
$pK_{In} = pH - \log \frac{[In^{-}]}{[HIn]}$
Rearranging the terms,we get:
$\log \frac{[In^{-}]}{[HIn]} = pH - pK_{In}$

(C) Methyl orange is a weak base indicator.
Its color change occurs in the acidic range.
The $pH$ range for the color change of methyl orange is approximately $3.1$ to $4.4$.
Therefore,the correct range is $3-5$.

(B) The correct answer is $(B)$.
Phenolphthalein is an acid-base indicator used for titrations involving strong bases and weak acids or strong bases and strong acids.
However,in the titration of ferrous sulphate $(FeSO_4)$ against potassium permanganate $(KMnO_4)$,the reaction is a redox titration.
$KMnO_4$ acts as a strong oxidizing agent and it oxidizes the phenolphthalein indicator,causing it to lose its characteristic color-changing property,making it unsuitable for this specific titration.

(C) Phenolphthalein is an acid-base indicator with a pH range of approximately $8.2$ to $10.0$.
It is suitable for titrations involving strong bases and weak acids or strong bases and strong acids.
In the case of $K_2CO_3$ $vs$ $HCl$,the titration involves a weak base $(K_2CO_3)$ and a strong acid $(HCl)$.
The equivalence point for this titration occurs in the acidic range (pH $< 7$),where phenolphthalein does not show a sharp color change.
Therefore,methyl orange is preferred for this titration.

(B) Phenolphthalein is a weak organic acid indicator with a $pH$ range of $8.2$ to $10.0$.
It is most suitable for titrations involving a strong base and a weak acid,as the equivalence point for such a titration lies in the basic range $(pH > 7)$.
Among the given options,$CH_3COOH$ is a weak acid and $NaOH$ is a strong base,making phenolphthalein the ideal indicator for this titration.

(D) The titration of $Na_2CO_3$ (a weak base) with $HCl$ (a strong acid) involves the formation of $NaHCO_3$ and then $NaCl$.
Methyl orange is the suitable indicator for this titration because its $pH$ range $(3.1-4.4)$ corresponds to the $pH$ at the equivalence point of the reaction between a strong acid and a weak base.

(C) In the titration of a strong acid and a strong base,the equivalence point occurs at $pH = 7$ (neutral).
Since both phenolphthalein (range $8.3 - 10.0$) and methyl orange (range $3.1 - 4.4$) show a sharp color change within the steep $pH$ change region of the titration curve (typically $pH \ 4$ to $10$),either indicator can be used.

(D) Phenolphthalein is the correct indicator for the titration of a weak acid with a strong base.
This is because the equivalence point of such a titration lies in the basic range ($pH$ range $8-10$).
Phenolphthalein changes its color from colourless to pink in this $pH$ range.

(A) Phenolphthalein is a weak organic acid represented as $HIn$. In an acidic medium,it exists in the unionized form which is colourless. In an alkaline medium,it dissociates to form the $In^{-}$ ion (the negative form),which is pink in colour. The $OH^{-}$ ions shift the equilibrium towards the formation of the $In^{-}$ ion.

(A) In the titration of $Na_2CO_3$ with $HCl$ using phenolphthalein,the reaction proceeds as: $Na_2CO_3 + HCl \rightarrow NaHCO_3 + NaCl$.
Phenolphthalein changes color in the $pH$ range of $8.2$ to $10.0$.
At the end point of this reaction,the solution contains $NaHCO_3$,which is weakly basic,but the phenolphthalein indicator becomes colourless as the $pH$ drops below $8.2$.
Therefore,the correct observation is that the solution becomes colourless.

(C) $Methyl$ $orange$ is an acid-base indicator that shows a red color in acidic solutions (pH range $3.1 - 4.4$).
Among the given options,$Hydrochloric$ $acid$ $(HCl)$ is an acid.
Therefore,$Methyl$ $orange$ will turn red in $Hydrochloric$ $acid$ solution.

(C) The titration of $Na_2CO_3$ (a weak base) with $HCl$ (a strong acid) results in the formation of $NaCl$,$H_2O$,and $CO_2$.
At the equivalence point,the solution contains dissolved $CO_2$ (carbonic acid),which makes the solution slightly acidic $(pH < 7)$.
Methyl red is an indicator that changes color in the acidic range ($pH$ range $4.2 - 6.3$),making it the most suitable choice for this titration.

(C) Starch solution is colourless,but it forms a deep blue-black complex in the presence of iodine.
Therefore,starch is commonly used as an indicator for the detection of traces of iodine in aqueous solutions.
Reaction: $\text{Starch} + \text{Iodine} \rightarrow \text{Starch-Iodine complex (blue-black colour)}$.

(D) $Methyl$ $orange$ is a common acid-base indicator.
In an acidic medium,it exhibits a red colour.
In an alkaline (basic) medium,it exhibits a yellow colour.
Therefore,the correct statement is that it gives a yellow colour in an alkaline medium and a red colour in an acid medium.

(C) The titration involves a weak base $(Na_2CO_3)$ and a strong acid $(HCl)$.
At the equivalence point,the resulting solution contains $NaCl$,$H_2O$,and $CO_2$ (which forms carbonic acid),making the solution slightly acidic.
Methyl red is an indicator that changes color in the acidic pH range $(4.2 - 6.3)$,making it the most suitable choice for this titration.

(B) Phenolphthalein is a weak organic acid with a $pH$ range of $8.2$ to $10.0$.
It is suitable for titrations involving a strong base because the equivalence point of such titrations falls within its $pH$ range.
In the case of $CH_3COOH$ (weak acid) and $NaOH$ (strong base),the resulting salt undergoes anionic hydrolysis,making the solution basic at the equivalence point $(pH > 7)$.
Therefore,phenolphthalein is the appropriate indicator for this titration.

(D) The titration of a weak acid (like oxalic acid) with a strong base (like $NaOH$) results in a salt of a weak acid and a strong base,which undergoes anionic hydrolysis to produce a basic solution at the equivalence point $(pH > 7)$.
Phenolphthalein is a suitable indicator for this titration because its $pH$ range $(8.2 - 10.0)$ coincides with the $pH$ change at the equivalence point of this specific acid-base reaction.

(B) Phenolphthalein is a weak organic acid indicator with a pH range of $8.2$ to $10.0$.
It is most suitable for titrations involving a strong base and a weak acid,as the equivalence point for such titrations lies in the basic range $(pH > 7)$.
In the given options,$CH_3COOH$ is a weak acid and $NaOH$ is a strong base.
Therefore,the titration of $CH_3COOH$ and $NaOH$ is best indicated by phenolphthalein.

(B) For a basic indicator,the equilibrium is represented as: $InOH \rightleftharpoons In^+ + OH^-$.
The dissociation constant is $K_b = \frac{[In^+][OH^-]}{[InOH]}$.
The color change occurs when $[In^+] = [InOH]$,which implies $K_b = [OH^-]$.
Given $K_b = 1 \times 10^{-11}$,we have $[OH^-] = 1 \times 10^{-11} \ M$.
Now,$pOH = -\log[OH^-] = -\log(1 \times 10^{-11}) = 11$.
Since $pH + pOH = 14$,we get $pH = 14 - 11 = 3$.

(D) Phenolphthalein is an organic compound with the formula $C_{20}H_{14}O_4$.
It acts as a weak organic acid in aqueous solution.
In its acidic form,it is colorless,while in its basic form,it dissociates to form a pink-colored ion.
Therefore,it is classified as a weak acid indicator.

(A) The $pH$ range of an indicator is the range over which it changes color.
Phenolphthalein is a weak acid indicator that changes color in the basic range,specifically between $pH$ $8.0$ and $9.8$.
Methyl orange changes color in the acidic range ($pH$ $3.1 - 4.4$).
Methyl red changes color in the acidic range ($pH$ $4.4 - 6.2$).
Litmus changes color in the near-neutral range ($pH$ $5.0 - 8.0$).
Therefore,the correct indicator for the $pH$ range $8 - 9.8$ is Phenolphthalein.

(A) For an acid-base indicator, the color change occurs when the concentration of the ionized form equals the concentration of the unionized form, i.e., $[In^-] = [HIn]$.
Using the Henderson-Hasselbalch equation for an indicator: $pH = pK_{in} + \log \frac{[In^-]}{[HIn]}$.
Since $[In^-] = [HIn]$, the ratio $\frac{[In^-]}{[HIn]} = 1$.
Therefore, $pH = pK_{in} + \log(1) = pK_{in}$.
Given $K_{in} = 1.0 \times 10^{-5}$, then $pK_{in} = -\log(1.0 \times 10^{-5}) = 5$.
Thus, $pH = 5$.

(D) In the titration of a weak acid (e.g.,$CH_3COOH$) and a strong base (e.g.,$NaOH$),the equivalence point lies in the basic range (pH > $7$).
Phenolphthalein is a suitable indicator for this titration because its pH range $(8.2 - 10.0)$ coincides with the steep rise in pH at the equivalence point of a weak acid-strong base titration.

(A) Phenolphthalein is a weak organic acid,represented as $HIn$. In an alkaline medium,the equilibrium $HIn \rightleftharpoons H^+ + In^-$ shifts to the right due to the removal of $H^+$ ions by $OH^-$ ions. The resulting anion $In^-$ is pink in color,which is responsible for the characteristic pink color of phenolphthalein in basic solutions.

(C) The titration of a weak acid $(CH_3COOH)$ with a strong base $(NaOH)$ results in a salt of a weak acid and a strong base,which undergoes anionic hydrolysis to form a basic solution at the equivalence point $(pH > 7)$.
Phenolphthalein is a suitable indicator for this titration because its $pH$ range $(8.2 - 10.0)$ coincides with the $pH$ change at the equivalence point of this specific acid-base reaction.

(A) Methyl orange is a pH indicator frequently used in titration. In an acidic medium (pH $< 3.1$),it exhibits a $Red$ color. In a basic medium (pH $> 4.4$),it exhibits a $Yellow$ color.

(C) Phenolphthalein is an acid-base indicator that changes color in the $pH$ range of $8.2$ to $10.0$.
It is suitable for titrations involving strong bases.
Option $C$ involves $KMnO_4$,which is an oxidizing agent,not a simple acid-base titration.
Furthermore,$KMnO_4$ acts as a self-indicator in redox titrations,making the use of phenolphthalein unnecessary and inappropriate for this specific reaction.

(D) The dissociation of an acid-base indicator $(HIn)$ can be represented as: $HIn \rightleftharpoons H^+ + In^-$.
Applying the law of mass action,the dissociation constant $K_{In}$ is given by: $K_{In} = \frac{[H^+][In^-]}{[HIn]}$.
Taking the negative logarithm on both sides: $-\log K_{In} = -\log [H^+] - \log \frac{[In^-]}{[HIn]}$.
This simplifies to: $pK_{In} = pH - \log \frac{[In^-]}{[HIn]}$.
Rearranging the terms gives: $\log \frac{[In^-]}{[HIn]} = pH - pK_{In}$.

(C) The titration of a strong base $(NaOH)$ and a weak acid (oxalic acid) results in a salt of a strong base and a weak acid,which undergoes anionic hydrolysis to produce a basic solution at the equivalence point $(pH > 7)$.
Phenolphthalein is a suitable indicator for this titration because its $pH$ range $(8.2 - 10.0)$ covers the equivalence point of this reaction.

(A) Methyl red is a synthetic acid-base indicator.
It changes its color in the $pH$ range of $4.2$ to $6.2$.
Below $pH$ $4.2$,it is red,and above $pH$ $6.2$,it is yellow.

(B) For the salt of a weak acid and strong base,the $pH$ is calculated as:
$pH = \frac{1}{2}(pK_w + pK_a(CH_3COOH) + \log C)$
Given $K_b(CH_3COO^-) = 10^{-9}$,so $K_a(CH_3COOH) = \frac{K_w}{K_b} = \frac{10^{-14}}{10^{-9}} = 10^{-5}$,which means $pK_a = 5$.
$pH = \frac{1}{2}(14 + 5 + \log 0.1) = \frac{1}{2}(19 - 1) = 9$.
For the indicator $HPh \rightleftharpoons H^+ + Ph^-$,the Henderson-Hasselbalch equation is:
$pH = pK_a(HPh) + \log \frac{[Ph^-]}{[HPh]}$
$9 = 9.6 + \log \frac{[Ph^-]}{[HPh]}$
$\log \frac{[Ph^-]}{[HPh]} = -0.6 = -2 \times 0.3 = -2 \log 2 = \log(2^{-2}) = \log(0.25)$.
So,$\frac{[Ph^-]}{[HPh]} = 0.25 = \frac{1}{4}$.
Let the total concentration be $[HPh] + [Ph^-] = 1$.
Since $[HPh] = 4[Ph^-]$,we have $4[Ph^-] + [Ph^-] = 1$,so $5[Ph^-] = 1$,$[Ph^-] = 0.2$.
Then $[HPh] = 1 - 0.2 = 0.8$.
The colourless form is $HPh$,so the fraction is $0.8$.

(B) Phenolphthalein is an acid-base indicator used to detect the equivalence point in acid-base titrations.
$FeSO_4 \, vs. \, KMnO_4$ is a redox titration,not an acid-base titration.
In a redox titration,the change in potential is monitored,and phenolphthalein does not change color based on the oxidation-reduction state of the reactants.
Therefore,it is not a suitable indicator for this reaction.

(A) In an alkaline medium,phenolphthalein undergoes a structural change where it loses a proton to form a dianion.
This dianion is a conjugated system that absorbs light in the visible region,resulting in the characteristic pink colour.
Therefore,the pink colour is due to the formation of the negative ion (dianion) of phenolphthalein.

(D) Calculate the milliequivalents $(meq)$ of each reactant:
$meq \text{ of } HCl = N \times V \text{ (in } mL) = 0.5 \times 20 = 10 \ meq$.
$meq \text{ of } NaOH = N \times V \text{ (in } mL) = 0.1 \times 35 = 3.5 \ meq$.
Since $meq \text{ of } HCl > meq \text{ of } NaOH$,the acid is in excess.
$meq \text{ of } HCl \text{ remaining} = 10 - 3.5 = 6.5 \ meq$.
Because the resulting solution contains excess acid,it is acidic.
Methyl orange is an indicator that turns orange-red in an acidic medium.

(D) In the titration of a weak acid and a strong base,the $pH$ at the equivalence point is greater than $7$ (typically between $8$ and $10$).
Phenolphthalein is the most suitable indicator because its $pH$ range ($8.2$ to $10$) coincides with the $pH$ change at the equivalence point of this titration.
Methyl orange and Methyl red are suitable for strong acid-strong base or strong acid-weak base titrations,as their $pH$ ranges are in the acidic region.

(A) In the titration of a strong acid and a weak base,the resulting salt undergoes cationic hydrolysis,making the solution acidic at the equivalence point $(pH < 7)$.
Methyl orange,which operates in the $pH$ range of $3.1$ to $4.4$,is the most suitable indicator for this titration as its color change interval matches the $pH$ at the equivalence point.

(B) The titration of $Na_2CO_3$ with $HCl$ involves the reaction of a weak base with a strong acid.
At the first equivalence point,$NaHCO_3$ is formed,and phenolphthalein can be used.
However,for the complete neutralization to $NaCl$,$CO_2$,and $H_2O$,the pH drops into the acidic range (around $3.7$ to $4.5$).
Methyl orange is the suitable indicator for this range,as it changes color in the acidic pH range.

(B) Statement $I$: In the titration of a strong acid with a weak base,the $pH$ at the equivalence point is in the acidic range $(pH < 7)$. Methyl orange has a $pH$ transition range of $3.2$ to $4.4$,which falls within the steep portion of the titration curve. Thus,it is a suitable indicator. Statement $I$ is true.
Statement $II$: In the titration of a weak acid (acetic acid) with a strong base $(NaOH)$,the $pH$ at the equivalence point is in the basic range $(pH > 7)$. Phenolphthalein has a $pH$ transition range of $8.2$ to $10.0$,which covers the steep portion of this titration curve. Thus,it is a suitable indicator. Statement $II$ is false.

(A) Methyl orange is an acid-base indicator that changes color depending on the $pH$ of the solution.
In an acidic medium (at the end point of a titration involving an acid),methyl orange undergoes protonation.
The structure changes from the benzenoid form (yellow color in basic medium) to the quinonoid form (red/pink color in acidic medium).
As shown in the reaction,the addition of $H^+$ ions leads to the formation of the quinonoid structure,which is responsible for the color change observed at the end point.

(D) Phenolphthalein $(HPh)$ is a weak organic acid.
In an acidic medium,the equilibrium $HPh \rightleftharpoons H^{+} + Ph^{-}$ shifts to the left due to the common ion effect of $H^{+}$ ions,keeping the concentration of the unionized form $(HPh)$ high,which is colourless.
In a basic medium,the $OH^{-}$ ions react with $H^{+}$ to form water,shifting the equilibrium to the right. This increases the concentration of the ionized form $(Ph^{-})$,which is pink in colour.
Therefore,Assertion $A$ is true,but Reason $R$ is false because phenolphthalein does dissociate in a basic medium.

(C) The neutralization of a weak acid $(CH_3COOH)$ with a strong base $(NaOH)$ results in a basic equivalence point $(pH > 7)$.
Phenolphthalein is a weak organic acid $(HIn)$ that is colorless in acidic medium and turns pink in basic medium due to the formation of its ionized form $(In^-)$.
Its $pH$ range is approximately $8.2 - 10.0$,which matches the equivalence point of the titration between a weak acid and a strong base.

(C) The titration involves a weak acid $(HA)$ and a strong base $(NaOH)$.
At the equivalence point,the salt $NaA$ undergoes hydrolysis: $A^- + H_2O \rightleftharpoons HA + OH^-$.
The $pH$ at the equivalence point is greater than $7$ (basic).
For a weak acid-strong base titration,the $pH$ range at the equivalence point typically falls between $8$ and $10$.
Phenolphthalein has a $pH$ range of $8.3$ to $11$,which covers the equivalence point $pH$ of this titration.
Therefore,Phenolphthalein is the most suitable indicator.

(A)

Indicator	$pH$ range
Methyl orange	$3.2-4.5$
Phenolphthalein	$8.3-10.5$

Indicators are chosen based on the $pH$ range of the equivalence point of the titration.
$1$. Methyl orange is suitable for titrations where the equivalence point falls in the acidic range $(pH \approx 3-5)$,such as strong acid vs. weak base.
$2$. Phenolphthalein is suitable for titrations where the equivalence point falls in the basic range $(pH \approx 8-10)$,such as weak acid vs. strong base.
$3$. Both indicators can be used for strong acid vs. strong base titrations because the $pH$ change at the equivalence point is very sharp $(pH \approx 4-10)$.
$4$. For weak acid vs. weak base titrations,there is no sharp change in $pH$ at the equivalence point,so no simple indicator is suitable. Therefore,the statement that Methyl orange may be used for a weak acid vs. weak base titration is incorrect.

(B) $pK_{In} = -\log(4 \times 10^{-10}) = -(\log 4 + \log 10^{-10}) = -(0.6 - 10) = 9.4$.
The indicator range is given by $pK_{In} \pm 1$,which is $8.4$ to $10.4$.
Statement $A$: Incorrect. Phenolphthalein is not suitable for weak acid-weak base titrations because the pH change is not sharp enough.
Statement $B$: Correct. The colour change range starts at $pH = pK_{In} - 1 = 9.4 - 1 = 8.4$.
Statement $C$: Incorrect. Phenolphthalein is a weak organic acid $(HIn)$.
Statement $D$: Correct. In an acidic medium,the equilibrium $HIn \rightleftharpoons H^+ + In^-$ shifts to the left,keeping it in the unionized $HIn$ form,which is colourless.
Therefore,statements $B$ and $D$ are correct. The total number of correct statements is $2$.

(C) Methyl orange is a weak base,not a weak acid. Therefore,Statement $I$ is $FALSE$.
The quinonoid form of methyl orange is red and is more intensely coloured than the benzenoid form,which is yellow. Therefore,Statement $II$ is $FALSE$.
Thus,both statements are incorrect.

(D)

Medium	Colour of Litmus
Acidic	Red
Neutral	Violet
Basic	Blue

Litmus is a natural indicator extracted from lichens. In an acidic solution,it turns red. In a neutral solution,it retains its original purple/violet colour. In a basic solution,it turns blue.