pH of strong Acids and strong Bases Questions in English

(A) The $pH$ of a solution is defined as $pH = -\log[H^+]$.
For a strong acid like $HCl$,the concentration of $H^+$ ions is equal to the concentration of the acid.
Given $[H^+] = 10 \ M$.
$pH = -\log(10) = -1$.
Since the $pH$ scale is typically defined for dilute solutions,for highly concentrated solutions,the $pH$ can be negative.
Therefore,the $pH$ of a $10 \ M$ $HCl$ solution is $-1$,which is less than $0$.

(C) The concentration of hydrogen ions is given as $[H^{+}] = 0.01 \, mole/litre = 10^{-2} \, mole/litre$.
Using the formula for $pH$,we have $pH = -\log[H^{+}]$.
Substituting the value,$pH = -\log(10^{-2})$.
Applying the logarithmic property $\log(a^b) = b \log(a)$,we get $pH = -(-2) \log(10)$.
Since $\log(10) = 1$,the $pH = 2$.

(C) For a strong base like $NaOH$,it dissociates completely as: $NaOH \rightarrow Na^+ + OH^-$.
Given concentration of $NaOH = 0.01 \, M = 10^{-2} \, M$.
Therefore,$[OH^-] = 10^{-2} \, M$.
$pOH = -\log[OH^-] = -\log(10^{-2}) = 2$.
Since $pH + pOH = 14$ at $25^{\circ}C$,we have $pH = 14 - pOH$.
$pH = 14 - 2 = 12$.

(B) $NaOH$ is a strong base,so it dissociates completely as: $NaOH \rightarrow Na^+ + OH^-$.
Given concentration of $NaOH = 0.001 \ M = 10^{-3} \ M$.
Therefore,$[OH^-] = 10^{-3} \ M$.
We know that the ionic product of water at $25^{\circ}C$ is $K_w = [H^+][OH^-] = 10^{-14}$.
$[H^+] = \frac{K_w}{[OH^-]} = \frac{10^{-14}}{10^{-3}} = 10^{-11} \ \text{mole/litre}$.

(B) The $pH$ of a solution is calculated using the formula: $pH = -\log[H^{+}]$.
Given $[H^{+}] = 3 \times 10^{-3} \ M$.
$pH = -\log(3 \times 10^{-3})$
$pH = -(\log 3 + \log 10^{-3})$
$pH = -(\log 3 - 3 \log 10)$
Since $\log 3 \approx 0.477$ and $\log 10 = 1$,
$pH = -(0.477 - 3) = -(-2.523) = 2.523$.
Therefore,the correct option is $B$.

(C) $KOH$ is a strong base and dissociates completely in water as $KOH \rightarrow K^+ + OH^-$.
Since the concentration of $KOH$ is $1 \ N$,the concentration of $[OH^-]$ is $1 \ M$.
$pOH = -\log[OH^-] = -\log(1) = 0$.
We know that $pH + pOH = 14$ at $25^{\circ}C$.
Therefore,$pH = 14 - pOH = 14 - 0 = 14$.

(B) $HCl$ is a strong acid and undergoes complete dissociation in aqueous solution.
$HCl \rightarrow H^{+} + Cl^{-}$
Since the concentration of $HCl$ is $0.01 \, M$,the concentration of $H^{+}$ ions produced will be equal to the concentration of $HCl$.
$[H^{+}] = 0.01 \, M = 10^{-2} \, M$.
Therefore,the correct option is $B$.

(B) $HCl$ is a strong acid and dissociates completely as $HCl \rightarrow H^{+} + Cl^{-}$.
Millimolar concentration means $[HCl] = 1 \times 10^{-3} \ M$.
Therefore,$[H^{+}] = 10^{-3} \ M$.
The $pH$ is calculated as $pH = -\log[H^{+}]$.
$pH = -\log(10^{-3}) = 3$.

(B) $Ba(OH)_2$ is a strong base that dissociates completely as follows:
$Ba(OH)_2 \to Ba^{2+} + 2OH^-$
Since one mole of $Ba(OH)_2$ produces two moles of $OH^-$,the concentration of $OH^-$ ions is:
$[OH^-] = 2 \times 0.05 \, M = 0.1 \, M$
Now,calculate the $pOH$:
$pOH = -\log[OH^-] = -\log(0.1) = 1$
Using the relation $pH + pOH = 14$ at $25^{\circ}C$:
$pH + 1 = 14$
$pH = 13$

(C) $H_2SO_4$ is a strong diprotic acid that dissociates completely as: $H_2SO_4 \rightarrow 2H^+ + SO_4^{2-}$.
Given concentration of $H_2SO_4 = 0.05 \, M$.
Therefore,the concentration of $H^+$ ions is $[H^+] = 2 \times 0.05 \, M = 0.1 \, M$.
The $pH$ is calculated as: $pH = -\log[H^+] = -\log(0.1) = 1$.

(A) For the first solution,$pH = 6$,so $[H^{+}]_1 = 10^{-6} \ M$.
For the second solution,$pH = 3$,so $[H^{+}]_2 = 10^{-3} \ M$.
When equal volumes $(V)$ are mixed,the total volume becomes $2V$.
The total concentration of $[H^{+}]$ is given by: $[H^{+}]_{total} = \frac{[H^{+}]_1 V + [H^{+}]_2 V}{2V} = \frac{10^{-6} + 10^{-3}}{2}$.
$[H^{+}]_{total} = \frac{0.001001}{2} = 5.005 \times 10^{-4} \ M$.
The $pH$ is calculated as: $pH = -\log(5.005 \times 10^{-4}) = 4 - \log(5.005) \approx 4 - 0.7 = 3.3$.

(A) Given the concentration of hydroxyl ions: $[OH^-] = 0.05 \ mol \ L^{-1} = 5 \times 10^{-2} \ M$.
Calculate $pOH$: $pOH = -\log[OH^-] = -\log(5 \times 10^{-2}) = 2 - \log 5 = 2 - 0.699 = 1.301$.
Calculate $pH$: $pH = 14 - pOH = 14 - 1.301 = 12.699$.
Since the $pH$ is greater than $7$,the solution is basic.

(A) Given $pH = 13$,we have $pOH = 14 - 13 = 1$.
Therefore,the concentration of hydroxide ions is $[OH^-] = 10^{-pOH} = 10^{-1} = 0.1 \ M$.
Since $Ca(OH)_2$ is a diacidic base,its molarity $M$ is related to normality $N$ by $N = 2 \times M$.
For $[OH^-] = 0.1 \ M$,the molarity of $Ca(OH)_2$ is $0.05 \ M$.
The molar mass of $Ca(OH)_2$ is $40 + 2 \times (16 + 1) = 74 \ g/mol$.
Weight $w = M \times \text{Molar Mass} \times \text{Volume in Liters} = 0.05 \times 74 \times 0.250 = 0.925 \ g$.

(B) $H_2SO_4$ is a strong diprotic acid that dissociates as: $H_2SO_4 + 2H_2O \rightarrow 2H_3O^{+} + SO_4^{2-}$.
For a $0.020 \ M \ H_2SO_4$ solution,the concentration of $H_3O^{+}$ is $[H_3O^{+}] = 2 \times 0.020 \ M = 0.040 \ M$.
In $2 \ L$ of solution,the moles of $H_3O^{+}$ are $n = M \times V = 0.040 \ mol/L \times 2 \ L = 0.080 \ mol$.
Thus,option $B$ is correct.

(B) $HCl$ is a strong acid that dissociates completely in aqueous solution as follows:
$HCl \rightarrow H^{+} + Cl^{-}$
Since the concentration of $HCl$ is $0.01 \, M$,the concentration of $H^{+}$ ions produced will also be $0.01 \, M$.
$0.01 \, M = 1 \times 10^{-2} \, M$.
Therefore,the concentration of $[H^{+}]$ is $10^{-2} \, M$.

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

(A) The strong electrolyte $M(OH)_2$ dissociates completely as: $M(OH)_2 \rightarrow M^{2+} + 2OH^-$.
The number of moles of $M(OH)_2 = 0.001 \ mol$.
The concentration of $M(OH)_2 = \frac{0.001 \ mol}{20 \ mL} \times 1000 \ mL/L = 0.05 \ M$.
Since $1 \ mol$ of $M(OH)_2$ gives $2 \ mol$ of $OH^-$,the concentration of $[OH^-] = 2 \times 0.05 \ M = 0.1 \ M$.
$pOH = -\log[OH^-] = -\log(0.1) = 1$.
$pH = 14 - pOH = 14 - 1 = 13$.

(B) The molar mass of $NaOH$ is $40 \ g/mol$.
Number of moles of $NaOH = \frac{4 \ g}{40 \ g/mol} = 0.1 \ mol$.
Volume of solution = $100 \ mL = 0.1 \ L$.
Concentration of $NaOH$ (or $[OH^{-}]$) = $\frac{0.1 \ mol}{0.1 \ L} = 1 \ M$.
Using the ionic product of water,$K_w = [H^{+}][OH^{-}] = 10^{-14}$ at $25 \ ^{\circ}C$.
$[H^{+}] = \frac{10^{-14}}{[OH^{-}]} = \frac{10^{-14}}{1} = 10^{-14} \ M$.

(B) For a very dilute solution of a strong acid,the contribution of $H^+$ ions from the auto-ionization of water cannot be neglected.
Let the total concentration of $[H^+] = x$.
The source of $H^+$ ions is from $HCl$ $(10^{-8} \, M)$ and water $(x - 10^{-8} \, M)$.
The concentration of $OH^-$ ions from water is also $(x - 10^{-8} \, M)$.
Using the ionic product of water: $K_W = [H^+][OH^-] = 10^{-14}$.
$x(x - 10^{-8}) = 10^{-14}$.
$x^2 - 10^{-8}x - 10^{-14} = 0$.
Using the quadratic formula $x = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{10^{-8} + \sqrt{(10^{-8})^2 - 4(1)(-10^{-14})}}{2}$.
$x = \frac{10^{-8} + \sqrt{10^{-16} + 4 \times 10^{-14}}}{2}$.
$x = \frac{10^{-8} + \sqrt{10^{-16} + 400 \times 10^{-16}}}{2} = \frac{10^{-8} + \sqrt{401 \times 10^{-16}}}{2}$.
$x = \frac{10^{-8} + 20.025 \times 10^{-8}}{2} = \frac{21.025 \times 10^{-8}}{2} = 1.05125 \times 10^{-7} \, M$.
Rounding to the nearest option,the correct answer is $1.0525 \times 10^{-7} \, M$.

(D) The molar mass of $NaOH$ is $40 \ g/mol$.
Number of moles of $NaOH = \frac{0.4 \ g}{40 \ g/mol} = 0.01 \ mol$.
Since the volume is $1 \ L$,the concentration $[NaOH] = 0.01 \ M = 10^{-2} \ M$.
$NaOH$ is a strong base,so $[OH^-] = 10^{-2} \ M$.
Using the relation $[H^+][OH^-] = 10^{-14}$,we get $[H^+] = \frac{10^{-14}}{10^{-2}} = 10^{-12} \ M$.
$pH = -\log[H^+] = -\log(10^{-12}) = 12$.

(B) For a strong base like $NaOH$,the concentration of $OH^{-}$ ions is equal to the concentration of the solution,i.e.,$[OH^{-}] = 0.1 \, M = 10^{-1} \, M$.
We know that the ionic product of water is $K_w = [H^{+}][OH^{-}] = 10^{-14}$ at $298 \, K$.
Substituting the value of $[OH^{-}]$,we get:
$[H^{+}] \times 10^{-1} = 10^{-14}$
$[H^{+}] = \frac{10^{-14}}{10^{-1}} = 10^{-13} \, M$.

(B) For a very dilute acid solution,the contribution of $H^+$ ions from the auto-ionization of water cannot be neglected.
Total $[H^+] = [H^+]_{HCl} + [H^+]_{H_2O}$.
Given $[H^+]_{HCl} = 10^{-8} \ M$.
Let the concentration of $H^+$ from water be $x$.
Then,$[H^+]_{total} = 10^{-8} + x$ and $[OH^-] = x$.
Since $K_w = [H^+][OH^-] = 10^{-14}$,we have $(10^{-8} + x)(x) = 10^{-14}$.
$x^2 + 10^{-8}x - 10^{-14} = 0$.
Solving this quadratic equation using the formula $x = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{-10^{-8} + \sqrt{(10^{-8})^2 - 4(1)(-10^{-14})}}{2} = \frac{-10^{-8} + \sqrt{10^{-16} + 4 \times 10^{-14}}}{2} \approx 0.95 \times 10^{-7} \ M$.
Total $[H^+] = 10^{-8} + 0.95 \times 10^{-7} = 1.05 \times 10^{-7} \ M$.
$pH = -\log[H^+] = -\log(1.05 \times 10^{-7}) = 7 - \log(1.05) \approx 6.98$.
Thus,the $pH$ lies between $6$ and $7$.

(B) The $pH$ of a solution is defined as $pH = -\log[H^+]$.
For the first solution,$pH = 3$,so $[H^+]_1 = 10^{-3} \, M$.
For the second solution,$pH = 3$,so $[H^+]_2 = 10^{-3} \, M$.
When two solutions of the same substance with the same $pH$ are mixed,the concentration of $[H^+]$ ions in the mixture remains the same.
$[H^+]_{mix} = \frac{V_1[H^+]_1 + V_2[H^+]_2}{V_1 + V_2} = \frac{100 \times 10^{-3} + 400 \times 10^{-3}}{100 + 400} = \frac{500 \times 10^{-3}}{500} = 10^{-3} \, M$.
Therefore,the $pH$ of the mixture is $-\log(10^{-3}) = 3$.

(B) For a strong acid like $HCl$,the concentration of $H^{+}$ ions is equal to the concentration of the acid.
$[H^{+}] = 10^{-5} \, M$
Using the ionic product of water at $25 \, ^\circ C$:
$[H^{+}] \times [OH^{-}] = 10^{-14}$
Substituting the value of $[H^{+}]$:
$10^{-5} \times [OH^{-}] = 10^{-14}$
$[OH^{-}] = \frac{10^{-14}}{10^{-5}} = 10^{-9} \, M$

(A) Given,$[OH^-] = 0.05 \ mol/L = 5 \times 10^{-2} \ mol/L$.
$pOH = -\log[OH^-] = -\log(5 \times 10^{-2}) = 2 - \log(5) = 2 - 0.699 = 1.301$.
Since $pH + pOH = 14$ at $25^{\circ}C$,$pH = 14 - 1.301 = 12.699$.
Since the $pH > 7$,the solution is basic.

(A) For a $\frac{N}{10} \ H_2SO_4$ solution,the normality $(N)$ is $0.1 \ N$.
Since $H_2SO_4$ is a strong acid,the concentration of $H^+$ ions is equal to the normality of the solution.
$[H^+] = 0.1 \ M = 10^{-1} \ M$.
The $pH$ is calculated as $pH = -\log[H^+]$.
$pH = -\log(10^{-1}) = 1$.

(C) The $pH$ of a solution is inversely proportional to the concentration of $H^+$ ions. Higher $[H^+]$ means lower $pH$,and lower $[H^+]$ means higher $pH$.
For a mixture of strong acids,the total $[H^+]$ is the sum of the concentrations of $H^+$ contributed by each acid.
$A: [H^+] = 2(M/10) + (M/20) + (M/10) = 0.2 + 0.05 + 0.1 = 0.35 \ M$
$B: [H^+] = 2(M/20) + (M/10) + (M/20) = 0.1 + 0.1 + 0.05 = 0.25 \ M$
$C: [H^+] = 2(M/20) + (M/10) + (M/40) = 0.1 + 0.1 + 0.025 = 0.225 \ M$
$D: [H^+] = 2(M/20) + (M/5) + (M/5) = 0.1 + 0.2 + 0.2 = 0.5 \ M$
Comparing the total $[H^+]$ values,mixture $C$ has the lowest concentration of $H^+$ ions $(0.225 \ M)$,therefore it will have the highest $pH$.

(D) The molar mass of $HCl$ is $36.5 \ g/mol$.
Number of moles of $HCl = \frac{1.825 \ g}{36.5 \ g/mol} = 0.05 \ mol$.
Concentration of $H^+ = [H^+] = \frac{\text{moles of } HCl}{\text{volume in } L} = \frac{0.05 \ mol}{5 \ L} = 0.01 \ M = 10^{-2} \ M$.
Using the ionic product of water,$K_w = [H^+][OH^-] = 10^{-14}$.
$[OH^-] = \frac{K_w}{[H^+]} = \frac{10^{-14}}{10^{-2}} = 10^{-12} \ M$.

(C) $pH$ is a measure of the hydrogen ion concentration,where $pH = -\log[H^+]$.
For basic solutions,a higher concentration of $OH^-$ ions corresponds to a higher $pH$.
$1 \ M \ NaOH$ is a strong base that dissociates completely to provide a high concentration of $OH^-$ ions,resulting in a $pH$ of $14$.
$1 \ M \ NH_3$ is a weak base with a lower $pH$ than $NaOH$.
Distilled water is neutral $(pH = 7)$.
Water saturated with chlorine is acidic due to the formation of $HCl$ and $HOCl$,resulting in a $pH < 7$.
Therefore,$1 \ M \ NaOH$ has the highest $pH$.

(B) For a very dilute acid solution ($10^{-8} \ M$ or less),the contribution of $H^+$ ions from the auto-ionization of water cannot be ignored.
Total $[H^+] = [H^+]_{HCl} + [H^+]_{H_2O}$.
Given $[H^+]_{HCl} = 10^{-9} \ M$.
From water,$[H^+]_{H_2O} \approx 10^{-7} \ M$ (at $25^{\circ}C$).
Total $[H^+] = 10^{-9} + 10^{-7} = 10^{-7} (0.01 + 1) = 1.01 \times 10^{-7} \ M$.
$pH = -\log[H^+] = -\log(1.01 \times 10^{-7}) = 7 - \log(1.01) \approx 7 - 0.004 = 6.996$.
Since $6.996$ is slightly less than $7$,the $pH$ lies between $6$ and $7$.

(C) For a $0.001 \, N \, KOH$ solution,the concentration of $[OH^-]$ is $10^{-3} \, M$.
$pOH = -\log[OH^-] = -\log(10^{-3}) = 3$.
Since $pH + pOH = 14$ at $25^{\circ}C$,
$pH = 14 - pOH = 14 - 3 = 11$.

(C) $1$. Calculate the number of moles of $NaOH$: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.2 \, g}{40 \, g/mol} = 0.005 \, mol$.
$2$. Calculate the molarity $(M)$ of the solution: $M = \frac{n}{V(L)} = \frac{0.005 \, mol}{0.1 \, L} = 0.05 \, M$.
$3$. Since $NaOH$ is a strong base,$[OH^-] = 0.05 \, M = 5 \times 10^{-2} \, M$.
$4$. Calculate $pOH$: $pOH = -\log[OH^-] = -\log(5 \times 10^{-2}) = 2 - \log(5) = 2 - 0.699 = 1.301$.
$5$. Calculate $pH$: $pH = 14 - pOH = 14 - 1.301 = 12.699$.

(C) Initial concentration of $HCl = 10^{-6} \ M$.
After $100$ times dilution,the new concentration of $HCl = \frac{10^{-6}}{100} = 10^{-8} \ M$.
In very dilute solutions,the contribution of $H^+$ ions from water cannot be ignored.
Total $[H^+] = [H^+]_{HCl} + [H^+]_{water} = 10^{-8} + 10^{-7} \ M$.
$[H^+] = 10^{-8} + 10 \times 10^{-8} = 11 \times 10^{-8} \ M$.
$pH = -\log[H^+] = -\log(11 \times 10^{-8}) = 8 - \log(11) = 8 - 1.0414 = 6.9586 \approx 6.95$.

(B) For a very dilute base solution like $10^{-10} \ M \ NaOH$,the contribution of $OH^-$ ions from the auto-ionization of water cannot be neglected.
The total $[OH^-] = [OH^-]_{\text{base}} + [OH^-]_{\text{water}} = 10^{-10} + 10^{-7} \approx 10^{-7} \ M$.
Thus,$pOH = -\log(10^{-7}) = 7$.
Since $pH + pOH = 14$,we get $pH = 14 - 7 = 7$.
Therefore,the $pH$ of the solution is approximately $7$.

(D) To have a $pH$ of $1.0$,the concentration of $[H^+]$ ions must be $0.1 \, M$.
For option $D$:
$n_{HCl} = M \times V = (1/5) \times 0.075 = 0.015 \, mol$.
$n_{NaOH} = M \times V = (1/5) \times 0.025 = 0.005 \, mol$.
Remaining $n_{HCl} = 0.015 - 0.005 = 0.010 \, mol$.
Total volume $= 75 \, mL + 25 \, mL = 100 \, mL = 0.1 \, L$.
$[H^+] = \frac{0.010 \, mol}{0.1 \, L} = 0.1 \, M$.
$pH = -\log[H^+] = -\log(0.1) = 1.0$.

(C) For a solution to have a $pH$ of $14$,the concentration of hydroxide ions $[OH^-]$ must be $1 \ M$.
For $1 \ N \ NaOH$,the normality is equal to the molarity since the $n$-factor is $1$.
Thus,$[OH^-] = 1 \ M$.
$pOH = -\log[OH^-] = -\log(1) = 0$.
Since $pH + pOH = 14$,we get $pH = 14 - 0 = 14$.

(C) To find the $pH$ of each solution:
$(1) 0.005 \, M \, H_2SO_4$: Since $H_2SO_4$ is a strong diprotic acid,$[H^+] = 2 \times 0.005 = 0.01 \, M = 10^{-2} \, M$. Thus,$pH = -\log(10^{-2}) = 2$.
$(2) 0.1 \, M \, Na_2SO_4$: This is a salt of a strong acid and a strong base,so it is neutral. $pH = 7$.
$(3) 10^{-2} \, M \, NaOH$: Since $NaOH$ is a strong base,$[OH^-] = 10^{-2} \, M$. Thus,$pOH = -\log(10^{-2}) = 2$,and $pH = 14 - 2 = 12$.
$(4) 0.01 \, M \, HCl$: Since $HCl$ is a strong monoprotic acid,$[H^+] = 0.01 \, M = 10^{-2} \, M$. Thus,$pH = -\log(10^{-2}) = 2$.
Comparing the $pH$ values,solutions $(1)$ and $(4)$ have the same $pH$ of $2$.

(D) $KOH$ is a strong base and dissociates completely as: $KOH \rightarrow K^+ + OH^-$.
Since the concentration of $KOH$ is $0.001 \ M$,the concentration of $[OH^-] = 0.001 \ M = 10^{-3} \ M$.
$pOH = -\log[OH^-] = -\log(10^{-3}) = 3$.
Using the relation $pH + pOH = 14$,we get $pH = 14 - 3 = 11$.

(D) For $pH = 1.0$,the concentration of $[H^{+}]$ must be $10^{-1} \ M = 0.1 \ M$.
Option $(d)$:
$M.eq. \ of \ HCl = 75 \ mL \times \frac{1}{5} \ M = 15 \ mmol$.
$M.eq. \ of \ NaOH = 25 \ mL \times \frac{1}{5} \ M = 5 \ mmol$.
Remaining $H^{+}$ ions $= 15 - 5 = 10 \ mmol$.
Total volume $= 75 + 25 = 100 \ mL$.
$[H^{+}] = \frac{10 \ mmol}{100 \ mL} = 0.1 \ M = 10^{-1} \ M$.
Therefore,$pH = -\log(10^{-1}) = 1.0$.

(C) The number of milliequivalents of $HCl$ is $N_1V_1 = 0.04 \times 100 = 4 \, meq$.
The number of milliequivalents of $NaOH$ is $N_2V_2 = 0.02 \times 100 = 2 \, meq$.
Since $HCl$ is a strong acid and $NaOH$ is a strong base,they neutralize each other. The remaining milliequivalents of $HCl$ are $4 - 2 = 2 \, meq$.
The total volume of the resulting solution is $100 \, mL + 100 \, mL = 200 \, mL$.
The normality of the resulting solution is $N_3 = \frac{2 \, meq}{200 \, mL} = 0.01 \, N = 10^{-2} \, N$.
Since $HCl$ is a strong monoprotic acid,$[H^+] = 10^{-2} \, M$.
The $pH$ is calculated as $pH = -\log [H^+] = -\log [10^{-2}] = 2$.

(B) The molar mass of $NaOH$ is $40 \, g/mol$.
The number of moles of $NaOH$ is $n = \frac{40 \times 10^{-3} \, g}{40 \, g/mol} = 10^{-3} \, mol$.
The concentration of $NaOH$ in $10 \, L$ is $[OH^-] = \frac{10^{-3} \, mol}{10 \, L} = 10^{-4} \, M$.
$pOH = -\log_{10}[OH^-] = -\log_{10}(10^{-4}) = 4$.
Using the relation $pH + pOH = 14$,we get $pH = 14 - 4 = 10$.

(D) Let the volume of each solution be $V \ L$.
Total volume of the mixture $= 2V \ L$.
Moles of $NaOH = 0.1 \times V = 0.1V$.
Moles of $HCl = 0.01 \times V = 0.01V$.
Since $NaOH$ is a strong base and $HCl$ is a strong acid,they react as: $NaOH + HCl \rightarrow NaCl + H_2O$.
Moles of $NaOH$ left unneutralized $= 0.1V - 0.01V = 0.09V$.
Concentration of $[OH^-] = \frac{0.09V}{2V} = 0.045 \ M$.
$pOH = -\log(0.045) = 1.3468 \approx 1.35$.
$pH = 14 - pOH = 14 - 1.35 = 12.65$.

(D) The $pH$ is defined as $pH = -\log[H^{+}]$,which implies $[H^{+}] = 10^{-pH}$.
For the three solutions:
$[H^{+}]_1 = 10^{-3} \ M$
$[H^{+}]_2 = 10^{-4} \ M$
$[H^{+}]_3 = 10^{-5} \ M$
Assuming equal volumes $V$ for each solution,the total volume of the mixture is $3V$.
The total moles of $H^{+}$ ions are $n_{total} = (10^{-3} \times V) + (10^{-4} \times V) + (10^{-5} \times V) = V(10^{-3} + 0.1 \times 10^{-3} + 0.01 \times 10^{-3}) = V(1.11 \times 10^{-3})$.
The final concentration is $[H^{+}]_{mix} = \frac{n_{total}}{3V} = \frac{1.11 \times 10^{-3}}{3} = 0.37 \times 10^{-3} \ M$.
Converting this to the form $\times 10^{-4} \ M$,we get $3.7 \times 10^{-4} \ M$.

(C) In an aqueous solution of $10^{-8} \ M \ HCl$,the total $[H^{+}]$ is the sum of $[H^{+}]$ from $HCl$ and $[H^{+}]$ from the dissociation of water.
Let $[H^{+}] = x$.
From $HCl$,$[H^{+}] = 10^{-8} \ M$.
From water,$[H^{+}] = [OH^{-}] = x - 10^{-8} \ M$.
Using the ionic product of water,$K_w = [H^{+}][OH^{-}] = 10^{-14}$.
$x(x - 10^{-8}) = 10^{-14}$
$x^{2} - 10^{-8}x - 10^{-14} = 0$
Solving this quadratic equation using the quadratic formula $x = \frac{-b + \sqrt{b^{2} - 4ac}}{2a}$:
$x = \frac{10^{-8} + \sqrt{(10^{-8})^{2} - 4(1)(-10^{-14})}}{2}$
$x = \frac{10^{-8} + \sqrt{10^{-16} + 4 \times 10^{-14}}}{2}$
$x = \frac{10^{-8} + \sqrt{10^{-16} + 400 \times 10^{-16}}}{2}$
$x = \frac{10^{-8} + \sqrt{401 \times 10^{-16}}}{2}$
$x = \frac{10^{-8} + 20.025 \times 10^{-8}}{2}$
$x = \frac{21.025 \times 10^{-8}}{2} = 1.05125 \times 10^{-7} \ M \approx 1.0525 \times 10^{-7} \ M$.