pH of strong Acids and strong Bases Questions in English

(N/A) For $HNO_3$,which is a strong monoprotic acid,$[H^+] = [HNO_3] = 0.002 \ M$.
$pH = -\log[H^+] = -\log(2 \times 10^{-3}) = 3 - \log 2 = 3 - 0.3010 = 2.699$.
$(b)$ For $H_2SO_4$,which is a strong diprotic acid,$[H^+] = 2 \times [H_2SO_4] = 2 \times 0.06 = 0.12 \ M$.
$pH = -\log[H^+] = -\log(0.12) = -\log(1.2 \times 10^{-1}) = 1 - \log 1.2 = 1 - 0.0792 = 0.9208$.

(A) $HNO_3$ is a strong acid,so $[H^+] = [HNO_3] = 0.001 \ M = 1.0 \times 10^{-3} \ M$.
$(b)$ $KOH$ is a strong base,so $[OH^-] = [KOH] = 0.0001 \ M = 1.0 \times 10^{-4} \ M$.
Using the ionic product of water,$K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$.
$[H^+] = \frac{K_w}{[OH^-]} = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-4}} = 1.0 \times 10^{-10} \ M$.

(A) $1$. Calculate the molar mass of $Ba(OH)_2$: $137 + 2 \times (16 + 1) = 137 + 34 = 171 \ g/mol$.
$2$. Calculate the number of moles of $Ba(OH)_2$: $n = \frac{0.837 \ g}{171 \ g/mol} = 0.00489 \ mol$.
$3$. Calculate the molarity of $Ba(OH)_2$: $M = \frac{0.00489 \ mol}{0.1 \ L} = 0.0489 \ M$.
$4$. Since $Ba(OH)_2$ is a strong base,$[OH^-] = 2 \times [Ba(OH)_2] = 2 \times 0.0489 = 0.0978 \ M$.
$5$. Calculate $pOH$: $pOH = -\log(0.0978) \approx 1.01$.
$6$. Calculate $pH$: $pH = 14 - pOH = 14 - 1.01 = 12.99 \approx 12.91$ (considering rounding differences in atomic mass precision).

(A) $1$. Calculate the molar mass of $Ca(OH)_2$: $40 + 2 \times (16 + 1) = 40 + 34 = 74 \ g/mol$.
$2$. Calculate the number of moles of $Ca(OH)_2$: $n = \frac{0.37 \ g}{74 \ g/mol} = 0.005 \ mol$.
$3$. Calculate the molarity of $Ca(OH)_2$: $M = \frac{0.005 \ mol}{0.2 \ L} = 0.025 \ M$.
$4$. Since $Ca(OH)_2$ is a strong base,it dissociates as $Ca(OH)_2 \rightarrow Ca^{2+} + 2OH^-$.
$5$. The concentration of $OH^-$ ions is $[OH^-] = 2 \times 0.025 \ M = 0.05 \ M$.
$6$. Calculate $pOH$: $pOH = -\log[OH^-] = -\log(0.05) = -\log(5 \times 10^{-2}) = 2 - \log(5) \approx 2 - 0.699 = 1.301$.
$7$. Calculate $pH$: $pH = 14 - pOH = 14 - 1.301 = 12.699 \approx 12.7$.

(B) For a strong base like $NaOH$,the concentration of $OH^-$ ions is $1.0 \times 10^{-8} \ M$.
However,we must also consider the contribution of $OH^-$ ions from the auto-ionization of water,which is $1.0 \times 10^{-7} \ M$.
Total $[OH^-] = 1.0 \times 10^{-8} + 1.0 \times 10^{-7} = 1.1 \times 10^{-7} \ M$.
$pOH = -\log[OH^-] = -\log(1.1 \times 10^{-7}) = 7 - \log(1.1) \approx 7 - 0.041 = 6.959$.
Since $pH + pOH = 14$,$pH = 14 - 6.959 = 7.041$.
Rounding to the nearest provided option,the value is $7.02$.

(D) The initial concentration of $H^+$ ions is $[H^+]_1 = 10^{-pH} = 10^{-5} \ M$.
Upon diluting the solution $100$ times,the new concentration becomes $[H^+]_2 = \frac{10^{-5}}{100} = 10^{-7} \ M$.
Since the concentration of $H^+$ from water $(10^{-7} \ M)$ cannot be ignored when the acid concentration is very low,the total $[H^+]$ is $[H^+]_{total} = [H^+]_{acid} + [H^+]_{water} = 10^{-7} + 10^{-7} = 2 \times 10^{-7} \ M$.
The $pH$ is calculated as $pH = -\log[H^+]_{total} = -\log(2 \times 10^{-7}) = 7 - \log 2 = 7 - 0.301 = 6.699 \approx 6.7$.

(B) Using the dilution formula $M_1V_1 = M_2V_2$:
$0.1 \ M \times 25 \ mL = M_2 \times 500 \ mL$
$M_2 = \frac{0.1 \times 25}{500} = 0.005 \ M$
Since $HCl$ is a strong acid,$[H^+] = [HCl] = 0.005 \ M = 5 \times 10^{-3} \ M$.
$pH = -\log[H^+] = -\log(5 \times 10^{-3}) = 3 - \log 5 = 3 - 0.698 = 2.302$.

(N/A) $1$. The $pH$ of the solution is $10.06$. Since $NaOH$ is a strong base,we first calculate the $pOH$ using the relation: $pH + pOH = 14$.
$2$. $pOH = 14 - 10.06 = 3.94$.
$3$. The concentration of hydroxide ions $[OH^-]$ is given by $[OH^-] = 10^{-pOH} = 10^{-3.94} \approx 1.148 \times 10^{-4} \ M$.
$4$. Since $NaOH$ dissociates completely as $NaOH \rightarrow Na^+ + OH^-$,the molarity of $NaOH$ is equal to $[OH^-]$,which is $1.148 \times 10^{-4} \ mol/L$.
$5$. The molar mass of $NaOH$ is $23 + 16 + 1 = 40 \ g/mol$.
$6$. Mass of $NaOH = \text{Molarity} \times \text{Volume} \times \text{Molar Mass} = 1.148 \times 10^{-4} \ mol/L \times 1 \ L \times 40 \ g/mol = 4.592 \times 10^{-3} \ g$.

(A) $1$. Initial moles of $Ba(OH)_2 = \text{Molarity} \times \text{Volume (L)} = 0.01 \ M \times 0.050 \ L = 5 \times 10^{-4} \ mol$.
$2$. Since $Ba(OH)_2$ is a strong base,it dissociates as $Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^-$.
$3$. Moles of $OH^- = 2 \times \text{moles of } Ba(OH)_2 = 2 \times 5 \times 10^{-4} = 10^{-3} \ mol$.
$4$. Total volume of the solution after adding water = $50 \ mL + 50 \ mL = 100 \ mL = 0.1 \ L$.
$5$. Concentration of $OH^- = \frac{10^{-3} \ mol}{0.1 \ L} = 0.01 \ M = 10^{-2} \ M$.
$6$. $pOH = -\log[OH^-] = -\log(10^{-2}) = 2$.
$7$. $pH = 14 - pOH = 14 - 2 = 12.0$.

(A) Step $1$: Calculate the total volume of the solution. $V_{total} = 50 \ mL + 50 \ mL = 100 \ mL = 0.1 \ L$.
Step $2$: Calculate the moles of $Ba(OH)_2$. $n = M \times V = 0.01 \ mol/L \times 0.05 \ L = 5 \times 10^{-4} \ mol$.
Step $3$: Calculate the concentration of $OH^-$ ions. Since $Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^-$,the moles of $OH^-$ are $2 \times (5 \times 10^{-4}) = 10^{-3} \ mol$.
$[OH^-] = \frac{10^{-3} \ mol}{0.1 \ L} = 0.01 \ M = 10^{-2} \ M$.
Step $4$: Calculate $pOH$. $pOH = -\log[OH^-] = -\log(10^{-2}) = 2$.
Step $5$: Calculate $pH$. $pH = 14 - pOH = 14 - 2 = 12.0$.

(A) $Ba(OH)_2$ is a strong base and dissociates completely as: $Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^-$.
The concentration of $OH^-$ ions is $[OH^-] = 2 \times [Ba(OH)_2] = 2 \times 0.0035 \ M = 0.007 \ M$.
Now,$pOH = -\log[OH^-] = -\log(0.007) = -\log(7 \times 10^{-3}) = 3 - \log(7) \approx 3 - 0.845 = 2.155$.
Since $pH + pOH = 14$,we have $pH = 14 - 2.155 = 11.845 \approx 11.84$.

(A) $1$. Given $pH = 2.25$ for $HCl$. The concentration of $H^+$ ions is $[H^+] = 10^{-pH} = 10^{-2.25}$.
$2$. $[H^+] = 10^{0.75} \times 10^{-3} \approx 5.623 \times 10^{-3} \ M$.
$3$. Since $HCl$ is a strong acid,$[HCl] = [H^+] = 5.623 \times 10^{-3} \ M$.
$4$. The neutralization reaction is $HCl + NaOH \rightarrow NaCl + H_2O$. The stoichiometry is $1:1$,so $n_{HCl} = n_{NaOH}$.
$5$. $n_{HCl} = M_{HCl} \times V_{HCl} = (5.623 \times 10^{-3} \ M) \times (0.300 \ L) = 1.687 \times 10^{-3} \ mol$.
$6$. $V_{NaOH} = \frac{n_{NaOH}}{M_{NaOH}} = \frac{1.687 \times 10^{-3} \ mol}{0.1 \ M} = 0.01687 \ L = 16.87 \ mL \approx 16.9 \ mL$.

(A) $1$. Calculate the millimoles of $H^+$ ions from $H_2SO_4$: $n(H^+) = 2 \times M \times V = 2 \times 0.05 \ M \times 300 \ mL = 30 \ mmol$.
$2$. Calculate the millimoles of $OH^-$ ions from $KOH$: $n(OH^-) = M \times V = 0.05 \ M \times 150 \ mL = 7.5 \ mmol$.
$3$. Calculate the remaining $H^+$ after neutralization: $n(H^+)_{rem} = 30 - 7.5 = 22.5 \ mmol$.
$4$. Calculate the total volume of the solution: $V_{total} = 300 \ mL + 150 \ mL = 450 \ mL$.
$5$. Calculate the concentration of $H^+$: $[H^+] = \frac{22.5 \ mmol}{450 \ mL} = 0.05 \ M$.
$6$. Calculate $pH$: $pH = -\log[H^+] = -\log(0.05) = -\log(5 \times 10^{-2}) = 2 - \log(5) = 2 - 0.699 = 1.301$.

(A) $H_2SO_4$ is a strong diprotic acid that dissociates completely in water as follows:
$H_2SO_4 \rightarrow 2H^+ + SO_4^{2-}$
Since the concentration of $H_2SO_4$ is $1 \times 10^{-4} \ M$,the concentration of $H^+$ ions is:
$[H^+] = 2 \times [H_2SO_4] = 2 \times 1 \times 10^{-4} \ M = 2 \times 10^{-4} \ M$
The $pH$ is calculated as:
$pH = -\log[H^+] = -\log(2 \times 10^{-4})$
$pH = -(\log 2 + \log 10^{-4}) = -(0.301 - 4) = 3.699 \approx 3.70$

(B) For $pH = 1$,$[H^+]_1 = 10^{-1} = 0.1 \ M$.
For $pH = 2$,$[H^+]_2 = 10^{-2} = 0.01 \ M$.
Using the dilution formula $M_1V_1 = M_2V_2$:
$0.1 \ M \times 1 \ L = 0.01 \ M \times V_2$.
$V_2 = \frac{0.1}{0.01} = 10 \ L$.
The volume of water to be added is $V_2 - V_1 = 10 \ L - 1 \ L = 9 \ L$.

(C) For the initial solution: $pH_1 = 1$,so $[H^+]_1 = 10^{-pH_1} = 10^{-1} \ M$.
Volume $V_1 = 1 \ L$.
Number of moles of $H^+ = M_1 \times V_1 = 10^{-1} \times 1 = 0.1 \ mol$.
For the final solution: $pH_2 = 2$,so $[H^+]_2 = 10^{-pH_2} = 10^{-2} \ M$.
Let the final volume be $V_2$.
Since the number of moles of $H^+$ remains constant: $M_1 \times V_1 = M_2 \times V_2$.
$0.1 \times 1 = 0.01 \times V_2$.
$V_2 = 10 \ L$.
Volume of water to be added = $V_2 - V_1 = 10 \ L - 1 \ L = 9 \ L$.

(B) Let the volume of each solution be $V \ L$.
Total volume of the mixture = $2V \ L$.
Moles of $H^+$ from $HCl$ = $0.01 \times V = 0.01V \ mol$.
Moles of $OH^-$ from $NaOH$ = $0.1 \times V = 0.1V \ mol$.
Since $OH^-$ is in excess,moles of $OH^-$ remaining = $0.1V - 0.01V = 0.09V \ mol$.
Concentration of $[OH^-]$ in the mixture = $\frac{0.09V}{2V} = 0.045 \ M$.
$pOH = -\log[OH^-] = -\log(0.045) = -(\log(4.5 \times 10^{-2})) = -(0.653 - 2) = 1.347$.
$pH = 14 - pOH = 14 - 1.347 = 12.653 \approx 12.65$.

(C) The titration involves a strong acid $(HCl)$ and a strong base $(NaOH)$.
Initially,the beaker contains $HCl$,which is a strong acid,so the initial $pH$ is low (around $1$).
As $NaOH$ is added,the $H^+$ ions are neutralized by $OH^-$ ions,causing the $pH$ to increase gradually.
Near the equivalence point,there is a very sharp or steep increase in $pH$ because the concentration of $H^+$ ions changes drastically.
After the equivalence point,the solution becomes basic due to excess $NaOH$,and the $pH$ levels off at a high value.
Graph $C$ correctly represents this characteristic sigmoidal curve for a strong acid-strong base titration,where the $pH$ rises sharply through $pH = 7$ at the equivalence point.

(B) $Ba(OH)_{2}$ is a strong base and dissociates completely as: $Ba(OH)_{2} \rightarrow Ba^{2+} + 2OH^{-}$.
Given concentration of $Ba(OH)_{2} = 0.005 \, M$.
Therefore,$[OH^{-}] = 2 \times 0.005 \, M = 0.01 \, M = 10^{-2} \, M$.
At $298 \, K$,the ionic product of water is $K_{w} = [H_{3}O^{+}][OH^{-}] = 10^{-14}$.
Substituting the value of $[OH^{-}]$:
$[H_{3}O^{+}] = \frac{10^{-14}}{10^{-2}} = 10^{-12} \, M$.
Thus,the concentration is $1 \times 10^{-12} \, mol \, L^{-1}$.

(D) $NaOH$ is a strong base,so it dissociates completely as $NaOH \rightarrow Na^{+} + OH^{-}$.
Given concentration of $NaOH = 0.001 \ M = 10^{-3} \ M$.
Therefore,$[OH^{-}] = 10^{-3} \ M$.
$pOH = -\log[OH^{-}] = -\log(10^{-3}) = 3$.
Since $pH + pOH = 14$ at $25^{\circ}C$,
$pH = 14 - pOH = 14 - 3 = 11$.

(B) The total moles of $H^{+}$ ions from $HCl$ is $n_{1} = M \times V = 0.01 \times 0.2 = 0.002 \ mol$.
The total moles of $H^{+}$ ions from $H_{2}SO_{4}$ is $n_{2} = M \times V \times 2 = 0.01 \times 0.4 \times 2 = 0.008 \ mol$.
Total moles of $H^{+} = 0.002 + 0.008 = 0.01 \ mol$.
Total volume of the mixture = $200 \ mL + 400 \ mL = 600 \ mL = 0.6 \ L$.
Concentration of $[H^{+}] = \frac{0.01 \ mol}{0.6 \ L} = \frac{1}{60} \ M$.
$pH = -\log[H^{+}] = -\log(\frac{1}{60}) = \log(60) = \log(6 \times 10) = \log(6) + \log(10) \approx 0.778 + 1 = 1.778 \approx 1.78$.

(A) The $pH$ values are determined as follows:
$(i)$ $0.1 \ M \ HCl$ is a strong acid,so its $pH = -\log[H^+] = -\log(0.1) = 1$.
(ii) $0.1 \ M \ KOH$ is a strong base,so its $pOH = -\log[OH^-] = 1$,which means $pH = 14 - 1 = 13$.
(iii) Tomato juice is acidic,typically having a $pH$ in the range of $4.0 - 4.5$.
(iv) Pure water is neutral,having a $pH = 7.0$ at $25^{\circ}C$.
Comparing these values: $1 (i) < 4.0-4.5 (iii) < 7.0 (iv) < 13 (ii)$.
Therefore,the correct order is $i < iii < iv < ii$.

(C) Given,
$pH_1 = 4$,so $[H^+]_1 = 10^{-4} \ M$.
Initial volume $V_1 = 10 \ mL$.
After dilution,$pH_2 = 5$,so $[H^+]_2 = 10^{-5} \ M$.
Using the dilution formula $M_1 V_1 = M_2 V_2$:
$10^{-4} \times 10 = 10^{-5} \times V_2$
$V_2 = \frac{10^{-4} \times 10}{10^{-5}} = 100 \ mL$.
Volume of water to be added = $V_2 - V_1 = 100 \ mL - 10 \ mL = 90 \ mL$.

(A) Given $pH = 12$,we know that $pOH = 14 - pH = 14 - 12 = 2$.
$[OH^-] = 10^{-pOH} = 10^{-2} \ M$.
Since $Ca(OH)_2$ dissociates completely as $Ca(OH)_2 \rightarrow Ca^{2+} + 2OH^-$,the concentration of $Ca(OH)_2$ is $[Ca(OH)_2] = \frac{[OH^-]}{2} = \frac{10^{-2}}{2} = 0.5 \times 10^{-2} = 5 \times 10^{-3} \ M$.
Number of millimoles $= \text{Molarity} \times \text{Volume in } mL = (5 \times 10^{-3} \ M) \times (100 \ mL) = 0.5 \ \text{millimoles}$.
This can be written as $5 \times 10^{-1} \ \text{millimoles}$.
Comparing with $x \times 10^{-1}$,we get $x = 5$.

(B) Total millimoles of $H^{+} = (600 \times 0.01) + (400 \times 0.01 \times 2) = 6 + 8 = 14 \, mmol$.
Total volume $= 600 + 400 = 1000 \, mL$.
Concentration of $[H^{+}] = \frac{14 \, mmol}{1000 \, mL} = 0.014 \, M = 1.4 \times 10^{-2} \, M$.
$pH = -\log[H^{+}] = -\log(1.4 \times 10^{-2}) = 2 - \log(1.4) = 2 - \log(14/10) = 2 - (\log 14 - \log 10) = 2 - (\log 2 + \log 7 - 1) = 2 - (0.30 + 0.84 - 1) = 2 - 0.14 = 1.86$.
Thus,$pH = 1.86 = 186 \times 10^{-2}$.

(B) For $pH = 1$,the concentration of $HCl$ is $[H^+]_1 = 10^{-pH} = 10^{-1} \ M = 0.1 \ M$.
For $pH = 2$,the concentration of $HCl$ is $[H^+]_2 = 10^{-pH} = 10^{-2} \ M = 0.01 \ M$.
Using the dilution formula $M_1 \times V_1 = M_2 \times V_2$:
$0.1 \ M \times 1 \ L = 0.01 \ M \times V_2$.
$V_2 = \frac{0.1}{0.01} \ L = 10 \ L$.
The volume of water to be added is $V_{added} = V_2 - V_1 = 10 \ L - 1 \ L = 9 \ L$.
Since $1 \ L = 1000 \ mL$,the volume of water added is $9 \times 1000 \ mL = 9000 \ mL$.

(A) Given $pH = 10.0$,so $pOH = 14 - 10 = 4.0$.
The concentration of hydroxide ions is $[OH^-] = 10^{-pOH} = 10^{-4} \ M$.
Since $Mg(OH)_2$ dissociates as $Mg(OH)_2 \rightarrow Mg^{2+} + 2OH^-$,the concentration of $Mg(OH)_2$ required is $[Mg(OH)_2] = \frac{[OH^-]}{2} = \frac{10^{-4}}{2} = 5 \times 10^{-5} \ M$.
For $1.0 \ L$ of solution,the number of moles of $Mg(OH)_2 = 5 \times 10^{-5} \ mol$.
Mass of $Mg(OH)_2 = \text{moles} \times \text{molar mass} = 5 \times 10^{-5} \ mol \times 58 \ g/mol = 290 \times 10^{-5} \ g = 2.9 \times 10^{-3} \ g$.
Converting to $mg$: $2.9 \times 10^{-3} \ g \times 1000 \ mg/g = 2.9 \ mg$.
The nearest integer value of $x$ is $3$.

(B) For an aqueous solution of $HCl$ with $pH = 1.0$,the concentration of hydrogen ions is $[H^{+}] = 10^{-pH} = 10^{-1} \ M = 0.1 \ M$.
When an equal volume of water is added,the total volume becomes double $(2V)$.
Since the number of moles of $HCl$ remains constant,the new concentration $[H^{+}]_{new}$ becomes half of the original concentration: $[H^{+}]_{new} = \frac{0.1 \ M}{2} = 0.05 \ M = 5 \times 10^{-2} \ M$.
The new $pH$ is calculated as: $pH = -\log[H^{+}]_{new} = -\log(5 \times 10^{-2}) = -(\log 5 + \log 10^{-2}) = -(\log(10/2) - 2) = -(1 - \log 2 - 2) = -(-1 - 0.30) = 1.3$.
Therefore,the $pH$ increases to $1.3$.

(C) Statement-$I$ is incorrect. For a very dilute solution of a strong acid like $HCl$ $(1.0 \times 10^{-8} \ M)$,the contribution of $H^{+}$ ions from the auto-ionization of water cannot be neglected.
The total $[H^{+}] = [H^{+}]_{HCl} + [H^{+}]_{water} = 10^{-8} + 10^{-7} \approx 1.1 \times 10^{-7} \ M$.
Therefore,$pH = -\log(1.1 \times 10^{-7}) \approx 6.96$,which is not $8$.
Statement-$II$ is correct. $HCl$ is a strong acid that dissociates completely in water,and the definition of $pH$ is indeed $-\log[H^{+}]$.

(C) For a very dilute solution of $HCl$ with concentration $10^{-8} \ M$,the contribution of $H^+$ ions from the auto-ionization of water cannot be neglected.
Total $[H^+] = [H^+]_{HCl} + [H^+]_{H_2O}$.
Since $[H^+]_{HCl} = 10^{-8} \ M$ and $[H^+]_{H_2O} \approx 10^{-7} \ M$,the total $[H^+]$ will be slightly greater than $10^{-7} \ M$.
Therefore,$pH = -\log[H^+] < -\log(10^{-7}) = 7$.
Thus,the $pH$ of a $10^{-8} \ M \ HCl$ solution is slightly less than $7$.

(D) Step $1$: Calculate the number of moles of $NaOH$. $n = \frac{\text{mass}}{\text{molar mass}} = \frac{4 \ g}{40 \ g \ mol^{-1}} = 0.1 \ mol$.
Step $2$: Calculate the molarity $(M)$ of the solution. $M = \frac{n}{V(L)} = \frac{0.1 \ mol}{0.5 \ L} = 0.2 \ M$.
Step $3$: Since $NaOH$ is a strong base,$[OH^-] = [NaOH] = 0.2 \ M$.
Step $4$: Calculate $pOH$. $pOH = -\log[OH^-] = -\log(0.2) = -(\log 2 - \log 10) = -(0.3010 - 1) = 0.6990$.
Step $5$: Calculate $pH$. $pH = 14 - pOH = 14 - 0.6990 = 13.3010$.

(B) $KOH$ is a strong base,so it dissociates completely: $[OH^-] = [KOH] = 0.002 \ M$.
$pOH = -\log_{10}[OH^-] = -\log_{10}(2 \times 10^{-3}) = 3 - \log_{10}(2) = 3 - 0.301 = 2.699 \approx 2.7$.
Using the relation $pH + pOH = 14$ at $298 \ K$:
$pH = 14 - 2.7 = 11.3$.

(A) The concentration of the solution is $1 \ mM = 10^{-3} \ M$.
Since $Ca(OH)_2$ is a strong base,it dissociates completely as: $Ca(OH)_2 \rightarrow Ca^{2+} + 2OH^-$.
For every $1 \text{ mole}$ of $Ca(OH)_2$,$2 \text{ moles}$ of $OH^-$ ions are produced.
Therefore,$[OH^-] = 2 \times 10^{-3} \ M$.
$pOH = -\log_{10} [OH^-] = -\log_{10} (2 \times 10^{-3})$.
$pOH = -(\log_{10} 2 + \log_{10} 10^{-3}) = -(0.301 - 3) = 2.699 \approx 2.7$.

(A) $H_2SO_4$ is a strong diprotic acid,so it dissociates completely as $H_2SO_4 \rightarrow 2H^+ + SO_4^{2-}$.
For a centimolar solution,concentration $C = 0.01 \ M = 10^{-2} \ M$.
The concentration of $H^+$ ions is $[H^+] = 2 \times C = 2 \times 10^{-2} \ M$.
$pH = -\log_{10} [H^+] = -\log_{10} (2 \times 10^{-2})$.
$pH = -(\log_{10} 2 + \log_{10} 10^{-2}) = -(\log_{10} 2 - 2) = 2 - \log_{10} 2$.
Using $\log_{10} 2 \approx 0.3010$,we get $pH = 2 - 0.3010 = 1.699 \approx 1.7$.

(D) For a strong monoacidic base,the concentration of hydroxide ions is $[OH^-] = 1 \times 10^{-4} \ M$.
$pOH = -\log_{10}[OH^-]$
$pOH = -\log_{10}(1 \times 10^{-4}) = 4$
Since $pH + pOH = 14$ at $25^{\circ} C$,
$pH = 14 - pOH = 14 - 4 = 10$.

(A) $H_2SO_4$ is a strong diprotic acid that dissociates completely in water:
$H_2SO_{4(aq)} + 2H_2O_{(l)} \longrightarrow 2H_3O^{+}_{(aq)} + SO_{4(aq)}^{2-}$
Since it is a strong acid,the concentration of hydronium ions is twice the concentration of the acid:
$[H_3O^{+}] = 2 \times c = 2 \times 0.01 \ M = 0.02 \ M = 2 \times 10^{-2} \ M$
The $pH$ is calculated as:
$pH = -\log_{10}[H_3O^{+}]$
$pH = -\log_{10}(2 \times 10^{-2})$
$pH = -(\log_{10}2 + \log_{10}10^{-2})$
$pH = -(0.3010 - 2)$
$pH = 2 - 0.3010 = 1.699$

(B) Perchloric acid $(HClO_4)$ is a strong monobasic acid,so it dissociates completely in water.
$[H_3O^{+}] = [HClO_4] = 1.36 \times 10^{-2} \ M$
$pH = -\log_{10}[H_3O^{+}]$
$pH = -\log_{10}(1.36 \times 10^{-2})$
$pH = -(\log_{10} 1.36 + \log_{10} 10^{-2})$
$pH = -\log_{10} 1.36 - (-2)$
$pH = 2 - 0.1335$
$pH = 1.8665 \approx 1.87$
Given the options,the closest value is $1.86$.

(C) $NaOH \longrightarrow Na^{+} + OH^{-}$
Since $NaOH$ is a strong base,$[OH^{-}] = [NaOH] = 0.005 \ M = 5 \times 10^{-3} \ M$.
$pOH = -\log_{10}[OH^{-}]$
$pOH = -\log_{10}(5 \times 10^{-3}) = -(\log_{10} 5 + \log_{10} 10^{-3})$
$pOH = -(0.699 - 3) = -(-2.301) = 2.301$.
We know that $pH + pOH = 14$ at $298 \ K$.
$pH = 14 - 2.301 = 11.699 \approx 11.7$.

(D) $pH = -\log_{10}[H^{+}]$
$= -\log_{10}(4.62 \times 10^{-4})$
$= -(\log_{10} 4.62 + \log_{10} 10^{-4})$
$= -(0.6646 - 4)$
$= 3.3354 \approx 3.34$

(D) For a strong dibasic acid,the concentration of hydronium ions is given by $[H_3O^{+}] = 2 \times c$.
$[H_3O^{+}] = 2 \times 0.01 \ M = 0.02 \ M = 2 \times 10^{-2} \ M$.
The $pH$ is calculated using the formula $pH = -\log_{10}[H_3O^{+}]$.
$pH = -\log_{10}(2 \times 10^{-2})$.
$pH = -(\log_{10} 2 + \log_{10} 10^{-2})$.
$pH = -0.3010 + 2$.
$pH = 1.699 \approx 1.7$.

(B) The formula for $pH$ is $pH = -\log [H^{+}]$.
Since the concentration of $H^{+}$ is very low $(< 10^{-7} \ M)$,the contribution of $H^{+}$ from the auto-ionization of water $(1.0 \times 10^{-7} \ M)$ must be considered.
$[H^{+}]_{total} = [H^{+}]_{acid} + [H^{+}]_{water} = 2.6 \times 10^{-8} + 1.0 \times 10^{-7} \ M$.
Converting to the same power of $10$: $[H^{+}]_{total} = 0.26 \times 10^{-7} + 1.0 \times 10^{-7} = 1.26 \times 10^{-7} \ M$.
Now,$pH = -\log(1.26 \times 10^{-7}) = 7 - \log(1.26)$.
Given $\log 2.6 = 0.4150$,we know $\log 1.26 = \log(2.6 / 2) = \log 2.6 - \log 2 = 0.4150 - 0.3010 \approx 0.114$.
$pH = 7 - 0.114 = 6.886 \approx 6.9$.

(B) $NaOH$ is a strong base and dissociates completely as $NaOH \rightarrow Na^+ + OH^-$.
Concentration of $NaOH = 1 \ mM = 1 \times 10^{-3} \ M$.
Therefore,$[OH^-] = 1 \times 10^{-3} \ M$.
$pOH = -\log[OH^-] = -\log(10^{-3}) = 3$.
Since $pH + pOH = 14$ at $25^{\circ}C$,
$pH = 14 - pOH = 14 - 3 = 11$.

(A) The $pH$ of a solution is calculated using the formula: $pH = -\log [H^{+}]$.
Given $[H^{+}] = 2.5 \times 10^{-3} \ mol \ dm^{-3}$.
Substituting the value: $pH = -\log (2.5 \times 10^{-3})$.
Using the property $\log (a \times b) = \log a + \log b$: $pH = -(\log 2.5 + \log 10^{-3})$.
$pH = -(\log 2.5 - 3)$.
$pH = 3 - \log 2.5$.
Given $\log 2.5 = 0.3979$,so $pH = 3 - 0.3979 = 2.6021$.
Rounding to one decimal place,the $pH$ is approximately $2.6$.

(D) $NaOH$ is a strong base,so it dissociates completely in water: $[NaOH] = [OH^-] = 0.02 \ M = 2 \times 10^{-2} \ M$.
$pOH = -\log[OH^-] = -\log(2 \times 10^{-2}) = 2 - \log 2 = 2 - 0.301 = 1.699 \approx 1.7$.
Using the relation $pH + pOH = 14$ at $25^{\circ}C$:
$pH = 14 - pOH = 14 - 1.7 = 12.3$.

(D) $H_2SO_4 \longrightarrow 2H^{+} + SO_4^{2-}$
$[H_2SO_4] = 0.005 \ M$
$[H^{+}] = 2 \times 0.005 = 0.01 \ M$
$pH = -\log[H^{+}] = -\log(0.01)$
$pH = 2$

(C) For a very dilute acid solution,the contribution of $H^+$ ions from the auto-ionization of water cannot be neglected.
The total concentration of $H^+$ ions is given by:
$[H^+]_{total} = [H^+]_{HCl} + [H^+]_{water} = 10^{-8} \ M + 10^{-7} \ M$.
$[H^+]_{total} = (0.1 \times 10^{-7} + 1 \times 10^{-7}) \ M = 1.1 \times 10^{-7} \ M$.
Now,calculate the $pH$:
$pH = -\log[H^+]_{total} = -\log(1.1 \times 10^{-7})$.
$pH = 7 - \log(1.1) \approx 7 - 0.0414 = 6.9586$.
Thus,the $pH$ lies between $6$ and $7$.

(A) The dissociation of $H_2SO_4$ is given by: $H_2SO_4 \rightarrow 2H^+ + SO_4^{2-}$.
Given $pH = 1$,we know that $[H^+] = 10^{-pH} = 10^{-1} = 0.1 \ M$.
Since $1 \ mol$ of $H_2SO_4$ produces $2 \ mol$ of $H^+$,the molarity $(M)$ of $H_2SO_4$ is $[H^+] / 2 = 0.1 / 2 = 0.05 \ M$.
Normality $(N)$ is calculated as $Molarity \times n$-factor.
For $H_2SO_4$,the $n$-factor is $2$.
Therefore,$N = 0.05 \times 2 = 0.1 \ N$.

(B) For a strong acid like $HCl$,the concentration of hydrogen ions is $[H^+] = 10^{-2} \ M$.
Using the ionic product of water,$K_w = [H^+][OH^-] = 10^{-14}$ at $298 \ K$.
Substituting the values: $[10^{-2}][OH^-] = 10^{-14}$.
Therefore,$[OH^-] = \frac{10^{-14}}{10^{-2}} = 10^{-12} \ mol \ dm^{-3}$.

(A) For an aqueous solution,the $pH$ scale ranges from $0$ to $14$. Acidic solutions have $pH < 7$,neutral solutions have $pH = 7$,and basic solutions have $pH > 7$.
Among the given options,$H_2SO_4$ and $HCl$ are strong acids,while $NaOH$ is a strong base. Therefore,the basic solutions will have a higher $pH$ than the acidic solutions.
Comparing the two basic solutions ($1 \ M \ NaOH$ and $0.1 \ M \ NaOH$):
For $1 \ M \ NaOH$: $[OH^-] = 1 \ M$. Thus,$pOH = -\log(1) = 0$. Since $pH + pOH = 14$,$pH = 14 - 0 = 14$.
For $0.1 \ M \ NaOH$: $[OH^-] = 0.1 \ M$. Thus,$pOH = -\log(0.1) = 1$. Since $pH + pOH = 14$,$pH = 14 - 1 = 13$.
Comparing the values,$1 \ M \ NaOH$ has the highest $pH$.

(B) Since $HCl$ is an acid,its $pH$ must be less than $7$.
For very dilute solutions of strong acids,the contribution of $H^+$ ions from the auto-ionization of water cannot be neglected.
From $HCl$,$[H^+]_{acid} = 10^{-8} \ M$.
From water,$[H^+]_{water} = 10^{-7} \ M$.
Total $[H^+] = 10^{-8} + 10^{-7} = 10^{-8}(1 + 10) = 11 \times 10^{-8} \ M$.
$pH = -\log[H^+] = -\log(11 \times 10^{-8}) = -(\log 11 + \log 10^{-8}) = -(1.0414 - 8) = 6.9586$.