pH of strong Acids and strong Bases Questions in English

(D) For the initial solution: $pH = 1$,so $[H^{+}]_1 = 10^{-1} = 0.1 \, M$.
For the final solution: $pH = 2$,so $[H^{+}]_2 = 10^{-2} = 0.01 \, M$.
Using the dilution formula $M_1 V_1 = M_2 V_2$:
$0.1 \, M \times 1 \, L = 0.01 \, M \times V_2$.
$V_2 = \frac{0.1}{0.01} = 10 \, L$.
The volume of water to be added is $V_2 - V_1 = 10 \, L - 1 \, L = 9 \, L$.

(C) Millimoles of $H^{+}$ from $HCl = 0.1 \times 10 = 1 \ mmol$.
Millimoles of $H^{+}$ from $H_2SO_4 = 0.2 \times 40 \times 2 = 16 \ mmol$.
Total millimoles of $H^{+} = 1 + 16 = 17 \ mmol$.
Total volume of solution = $10 \ mL + 40 \ mL = 50 \ mL$.
Concentration of $H^{+} = \frac{17 \ mmol}{50 \ mL} = 0.34 \ M$.
$pH = -\log [H^{+}] = -\log (0.34) = -\log (\frac{34}{100}) = -\log (\frac{17}{50}) = -(\log 17 - \log 50) = -(\log 17 - (\log 5 + \log 10)) = -(1.23 - (0.7 + 1)) = -(1.23 - 1.7) = 0.47$.

(B) Initially,the $NaCl$ solution is neutral,so its $pH = 7$.
After adding $1 \ mL$ of $1 \ M \ HCl$ to $99 \ mL$ of $NaCl$ solution,the total volume becomes $100 \ mL$.
The concentration of $H^+$ ions in the new solution is $[H^+] = \frac{n}{V} = \frac{1 \ mL \times 1 \ M}{100 \ mL} = 0.01 \ M = 10^{-2} \ M$.
The new $pH$ of the solution is $pH = -\log[H^+] = -\log(10^{-2}) = 2$.
The change in $pH$ is $X = |pH_{initial} - pH_{final}| = |7 - 2| = 5$ units.

(C) Given $pH = 14$.
Since $pH + pOH = 14$,we have $pOH = 14 - 14 = 0$.
$pOH = -\log[OH^-] = 0$,which implies $[OH^-] = 10^0 = 1 \ M$.
$Ca(OH)_2$ undergoes complete ionization as $Ca(OH)_2 \rightarrow Ca^{2+} + 2OH^-$.
Thus,$1 \ mole$ of $Ca(OH)_2$ produces $2 \ moles$ of $OH^-$.
Therefore,the concentration of $Ca(OH)_2$ required is $[Ca(OH)_2] = \frac{[OH^-]}{2} = \frac{1}{2} = 0.5 \ M$.
For $250 \ mL$ $(0.25 \ L)$ of solution,the number of moles required is $n = M \times V(L) = 0.5 \ mol \ L^{-1} \times 0.25 \ L = 0.125 \ mole$.

(B) Milli-equivalents of $HCl = 25 \times 0.08 = 2.0 \ mmol$.
Milli-equivalents of $NaOH = 25 \times 0.1 = 2.5 \ mmol$.
Since $NaOH$ is in excess,the remaining $OH^{-}$ concentration is calculated as:
$[OH^{-}] = \frac{2.5 - 2.0}{500 \ mL} = \frac{0.5}{500} = 10^{-3} \ M$.
Now,$pOH = -\log[OH^{-}] = -\log(10^{-3}) = 3$.
Since $pH + pOH = 14$,we have $pH = 14 - 3 = 11$.

(D) Using the formula $pH = -\log [H^{+}]$.
Initial $pH = 1$,so $[H^{+}]_{1} = 10^{-1} = 0.1 \ M$.
Final $pH = 2$,so $[H^{+}]_{2} = 10^{-2} = 0.01 \ M$.
For dilution,the number of moles of $HCl$ remains constant: $M_{1}V_{1} = M_{2}V_{2}$.
Substituting the values: $0.1 \ M \times 1 \ L = 0.01 \ M \times V_{2}$.
$V_{2} = \frac{0.1}{0.01} = 10 \ L$.
The volume of water to be added is $V_{2} - V_{1} = 10 \ L - 1 \ L = 9 \ L$.

(C) The relationship between $pH$ and $H^{+}$ concentration is given by the formula: $pH = -\log[H^{+}]$.
Initially,$pH_1 = 5.0$,so $[H^{+}]_1 = 10^{-5} \ M$.
Finally,$pH_2 = 2.0$,so $[H^{+}]_2 = 10^{-2} \ M$.
The ratio of the final concentration to the initial concentration is: $\frac{[H^{+}]_2}{[H^{+}]_1} = \frac{10^{-2}}{10^{-5}} = 10^{3} = 1000$.
Therefore,the concentration of $H^{+}$ ions becomes $1000$ times the initial concentration.

(D) $HClO_4$ is a strong acid,so it will have a very low $pH$.
$CH_3COOH$ is a weak acid,which will have a $pH$ lower than $7$.
$HCl$ is a strong acid,so it will have a low $pH$.
$NaCl$ is a salt formed from a strong acid $(HCl)$ and a strong base $(NaOH)$,so it undergoes no hydrolysis and remains neutral with a $pH = 7$.
Since the other three solutions are acidic $(pH < 7)$,the $0.2 \, M \, NaCl$ solution has the highest $pH$.

(D) For a strong acid like $HBr$,the concentration of $H^+$ ions is given by $[H^+] = 10^{-pH}$.
Given $pH = 4$,so $[H^+] = 10^{-4} \ M$.
Since $HBr$ is a monoprotic acid,the normality $(N)$ is equal to the molarity $(M)$,so $N = 10^{-4} \ N$.
The volume of the solution is $V = 100 \ mL = 0.1 \ L$.
The number of equivalents is calculated as $\text{Equivalents} = N \times V \text{ (in liters)}$.
$\text{Equivalents} = 10^{-4} \times 0.1 = 10^{-5} \ eq$.

(D) For $0.1 \, M \, H_2SO_4$,the concentration of $H^+$ ions is $[H^+] = 2 \times 0.1 = 0.2 \, M$.
$pH_1 = -\log(0.2) = -(\log 2 + \log 10^{-1}) = -0.3 + 1 = 0.7$.
For $0.1 \, N \, H_2SO_4$,the normality is given as $0.1 \, N$,which means $[H^+] = 0.1 \, M$.
$pH_2 = -\log(0.1) = 1$.
The ratio of $pH$ is $pH_1 : pH_2 = 0.7 : 1 = 7 : 10$.

(B) For $pH = 1$,the concentration of $[H^{+}]$ must be $10^{-1} \ M$ or $0.1 \ M$ $(M/10)$.
For option $B$: $75 \ mL \ \frac{M}{5} \ HCl + 25 \ mL \ \frac{M}{5} \ NaOH$.
$25 \ mL \ \frac{M}{5} \ NaOH$ will neutralize $25 \ mL \ \frac{M}{5} \ HCl$.
Remaining $HCl$ volume = $75 \ mL - 25 \ mL = 50 \ mL$ of $\frac{M}{5} \ HCl$.
Total volume of the solution = $75 \ mL + 25 \ mL = 100 \ mL$.
New concentration of $HCl = \frac{M}{5} \times \frac{50 \ mL}{100 \ mL} = \frac{M}{10} = 0.1 \ M$.
Since $HCl$ is a strong acid,$[H^{+}] = 0.1 \ M$.
$pH = -\log_{10}[H^{+}] = -\log_{10}(10^{-1}) = 1$.

(D) Given that $NaOH$ is a strong base,it dissociates completely as: $NaOH \rightarrow Na^+ + OH^-$.
Therefore,$[OH^-] = [NaOH] = 5.0 \times 10^{-2} \ M$.
Now,calculate $pOH$:
$pOH = -\log [OH^-] = -\log (5.0 \times 10^{-2})$
$pOH = -(\log 5 + \log 10^{-2}) = -(\log (10/2) - 2)$
$pOH = -(1 - \log 2 - 2) = -(-1 - 0.3) = 1.30$.
Using the relation $pH + pOH = 14$ at $25^{\circ}C$:
$pH = 14 - pOH = 14 - 1.30 = 12.70$.

(D) Given at $330 \ K$,$K_w = 10^{-13.6}$.
We know that $pK_w = pH + pOH$.
Since $pK_w = - \log K_w$,we have $pK_w = 13.6$.
Given $[OH^{-}] = 10^{-4} \ M$,so $pOH = - \log [OH^{-}] = - \log 10^{-4} = 4$.
Using the relation $pH + pOH = pK_w$:
$pH + 4 = 13.6$
$pH = 13.6 - 4 = 9.6$.

(B) For a very dilute acid solution,the contribution of $H^{+}$ ions from the auto-ionization of water cannot be neglected.
Given concentration of $HCl = 10^{-9} \ M$.
Total $[H^{+}] = [H^{+}]_{HCl} + [H^{+}]_{H_2O}$.
$[H^{+}] = 10^{-9} + 10^{-7} \ M$ (since $[H^{+}]$ from water is $10^{-7} \ M$ at $25^{\circ}C$).
$[H^{+}] = 0.01 \times 10^{-7} + 10^{-7} = 1.01 \times 10^{-7} \ M$.
$pH = -\log [H^{+}] = -\log (1.01 \times 10^{-7})$.
$pH = 7 - \log (1.01) \approx 7 - 0.0043 = 6.9957$.
Thus,the $pH$ is between $6$ and $7$.

(C) Initial concentration of $HCl = 10^{-6} \ M$.
After diluting $100$ times,the new concentration of $HCl$ is $10^{-6} / 100 = 10^{-8} \ M$.
Since the concentration is very low,we must consider the contribution of $H^{+}$ ions from water.
Total $[H^{+}] = [H^{+}]_{HCl} + [H^{+}]_{H_2O} = 10^{-8} + 10^{-7} = 1.1 \times 10^{-7} \ M$.
$pH = -\log[H^{+}] = -\log(1.1 \times 10^{-7}) = 7 - \log(1.1) \approx 7 - 0.04 = 6.96$.

(A) $HCl$ is a strong acid and it dissociates completely in aqueous solution as follows:
$HCl(aq) \rightarrow H^{+}(aq) + Cl^{-}(aq)$
Since $HCl$ is a strong electrolyte,the concentration of $H^{+}$ ions produced is equal to the initial concentration of the $HCl$ solution.
Given concentration of $HCl = 10^{-4} \ M$.
Therefore,$[H^{+}] = 10^{-4} \ M$.

(B) For $HCl$ solution: $pH = 2$,so $[H^{+}] = 10^{-2} \ M$. Milli-moles of $H^{+} = 100 \ mL \times 10^{-2} \ M = 1 \ mmol$.
For $NaOH$ solution: $pH = 12$,so $pOH = 14 - 12 = 2$. Thus,$[OH^{-}] = 10^{-2} \ M$. Milli-moles of $OH^{-} = 200 \ mL \times 10^{-2} \ M = 2 \ mmol$.
Reaction: $H^{+} + OH^{-} \longrightarrow H_{2}O$.
Remaining milli-moles of $OH^{-} = 2 - 1 = 1 \ mmol$.
Total volume of the mixture = $100 \ mL + 200 \ mL = 300 \ mL$.
Concentration of $[OH^{-}] = \frac{1 \ mmol}{300 \ mL} = 3.33 \times 10^{-3} \ M$.
$pOH = -\log(3.33 \times 10^{-3}) = 3 - \log(3.33) \approx 3 - 0.52 = 2.48$.
$pH = 14 - pOH = 14 - 2.48 = 11.52$.

(A) Given $pH = 12$,so $pOH = 14 - 12 = 2$.
Concentration of hydroxide ions: $[OH^-] = 10^{-pOH} = 10^{-2} \ M$.
The dissociation reaction is: $Mg(OH)_2 \rightleftharpoons Mg^{2+} + 2OH^-$.
Since $1 \ mol$ of $Mg(OH)_2$ produces $2 \ mol$ of $OH^-$,the concentration of $Mg(OH)_2$ is $[Mg(OH)_2] = \frac{[OH^-]}{2} = \frac{10^{-2}}{2} = 0.5 \times 10^{-2} \ M$.
Molar mass of $Mg(OH)_2 = 24 + 2 \times (16 + 1) = 58 \ g/mol$.
Using the formula $Molarity = \frac{w_B \times 1000}{M_B \times V(mL)}$:
$0.5 \times 10^{-2} = \frac{w_B \times 1000}{58 \times 200}$.
$w_B = 0.5 \times 10^{-2} \times 58 \times 0.2 = 0.58 \ g$.

(B) First,calculate the number of moles of $NaOH$: $\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{4 \ g}{40 \ g/mol} = 0.1 \ mol$.
Next,calculate the molarity of the $NaOH$ solution: $[OH^{-}] = \frac{0.1 \ mol}{1 \ L} = 0.1 \ M = 10^{-1} \ M$.
Calculate $pOH$: $pOH = -\log[OH^{-}] = -\log(10^{-1}) = 1$.
Calculate $pH$: $pH = 14 - pOH = 14 - 1 = 13$.
Finally,calculate the $H^{+}$ ion concentration: $[H^{+}] = 10^{-pH} = 10^{-13} \ M$.

(B) Given $pH = 5$.
Since $pH = -\log[H^+]$,we have $[H^+] = 10^{-5} \ M$.
For $H_2SO_4$,which is a strong diprotic acid,the normality $N$ is equal to the molarity $M$ multiplied by the $n$-factor $(n=2)$.
However,$[H^+] = N$ for a strong acid solution.
Thus,$N = [H^+] = 10^{-5} \ N$.
Volume of solution $V = 100 \ mL = 0.1 \ L$.
Number of equivalents = $N \times V(L) = 10^{-5} \times 0.1 = 10^{-6} \ \text{equivalents}$.

(B) The total volume of the resulting solution is $1 \, mL + 999 \, mL = 1000 \, mL = 1 \, L$.
The number of moles of $H^+$ ions provided by $HCl$ is $n_{H^+} = M \times V(L) = 0.1 \, mol/L \times 0.001 \, L = 10^{-4} \, mol$.
The concentration of $H^+$ ions in the final solution is $[H^+] = \frac{n_{H^+}}{V_{total}} = \frac{10^{-4} \, mol}{1 \, L} = 10^{-4} \, M$.
The $pH$ of the solution is calculated as $pH = -\log[H^+] = -\log(10^{-4}) = 4$.

(C) Let the volume of each solution be $V$.
For the first solution,$[H^{+}]_1 = 10^{-3} \, M$.
For the second solution,$[H^{+}]_2 = 10^{-5} \, M$.
The total number of moles of $H^{+}$ ions is $n = (10^{-3} \times V) + (10^{-5} \times V) = V(10^{-3} + 0.01 \times 10^{-3}) = 1.01 \times 10^{-3} \times V$.
The total volume of the mixture is $2V$.
The resultant concentration of $H^{+}$ ions is $[H^{+}]_{mix} = \frac{1.01 \times 10^{-3} \times V}{2V} = 0.505 \times 10^{-3} = 5.05 \times 10^{-4} \, M$.
The resultant $pH$ is calculated as $pH = -\log[H^{+}]_{mix} = -\log(5.05 \times 10^{-4}) = 4 - \log(5.05) \approx 4 - 0.703 = 3.297 \approx 3.3$.

(C) $Ba(OH)_2$ is a strong base and dissociates as: $Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^-$.
Since the concentration of $Ba(OH)_2$ is $10^{-8} \ M$,the concentration of $OH^-$ ions from the base is $[OH^-]_{base} = 2 \times 10^{-8} \ M$.
At such low concentrations,the contribution of $OH^-$ from water $(10^{-7} \ M)$ cannot be ignored.
Total $[OH^-] = [OH^-]_{base} + [OH^-]_{water} = 2 \times 10^{-8} + 10^{-7} = 1.2 \times 10^{-7} \ M$.
Now,$pOH = -\log[OH^-] = -\log(1.2 \times 10^{-7}) = 7 - \log(1.2) \approx 7 - 0.079 = 6.921$.
Since $pH + pOH = 14$,$pH = 14 - 6.921 = 7.079 \approx 7.08$.

(B) Step $1$: Calculate the molar mass of $NaOH = 23 + 16 + 1 = 40 \ g/mol$.
Step $2$: Calculate the number of moles of $NaOH = \frac{\text{mass}}{\text{molar mass}} = \frac{40 \times 10^{-3} \ g}{40 \ g/mol} = 10^{-3} \ mol$.
Step $3$: Calculate the concentration of $OH^-$ ions: $[OH^-] = \frac{\text{moles}}{\text{volume in L}} = \frac{10^{-3} \ mol}{10 \ L} = 10^{-4} \ M$.
Step $4$: Calculate $pOH = -\log[OH^-] = -\log(10^{-4}) = 4$.
Step $5$: Calculate $pH = 14 - pOH = 14 - 4 = 10$.

(A) $Ba(OH)_2$ is a strong base and dissociates completely as: $Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^-$.
Since the concentration of $Ba(OH)_2$ is $0.005 \ M$,the concentration of $OH^-$ ions is $[OH^-] = 2 \times 0.005 \ M = 0.01 \ M = 10^{-2} \ M$.
Now,calculate $pOH$: $pOH = -\log[OH^-] = -\log(10^{-2}) = 2$.
Finally,calculate $pH$ using the relation: $pH + pOH = 14$.
$pH = 14 - 2 = 12$.

(C) For a strong acid,the $pH$ is given by $pH = -\log[H^+]$.
Initially,$pH_1 = 1$,so $[H^+]_1 = 10^{-1} \ M = 0.1 \ M$.
Initial volume $V_1 = 100 \ mL$.
We want the final $pH_2 = 2$,so $[H^+]_2 = 10^{-2} \ M = 0.01 \ M$.
Using the dilution formula $M_1V_1 = M_2V_2$:
$0.1 \ M \times 100 \ mL = 0.01 \ M \times V_2$.
$V_2 = (0.1 \times 100) / 0.01 = 1000 \ mL$.
The volume of water to be added is $V_2 - V_1 = 1000 \ mL - 100 \ mL = 900 \ mL$.

(B) For a very dilute acid solution,the contribution of $H^+$ ions from the auto-ionization of water cannot be ignored.
Total $[H^+] = [H^+]_{HCl} + [H^+]_{H_2O}$.
Given $[H^+]_{HCl} = 10^{-8} \ M$.
Let the concentration of $H^+$ from water be $x$.
Then $[H^+]_{total} = 10^{-8} + x$.
Since the water equilibrium is $[H^+][OH^-] = 10^{-14}$ and $[H^+] = [OH^-] + 10^{-8}$,we solve the quadratic equation $x^2 + 10^{-8}x - 10^{-14} = 0$.
Solving this gives $[H^+] \approx 1.05 \times 10^{-7} \ M$.
$pH = -\log(1.05 \times 10^{-7}) \approx 6.98$.
Thus,the $pH$ is slightly less than $7$.

(D) To determine the maximum $pH$,we need to find the solution with the highest concentration of hydroxide ions $[OH^-]$.
$A$. $10^{-3} \ M \ NH_4OH$: This is a weak base. $[OH^-] = \sqrt{K_b \times C} < 10^{-3} \ M$.
$B$. $10^{-1} \ M \ NH_4OH$: This is a weak base. $[OH^-] = \sqrt{K_b \times 10^{-1}} \approx 10^{-3} \ M$ (assuming $K_b \approx 10^{-5}$).
$C$. $10^{-3} \ M \ NaOH$: This is a strong base. $[OH^-] = 10^{-3} \ M$.
$D$. $10^{-3} \ M \ Ba(OH)_2$: This is a strong base. $[OH^-] = 2 \times 10^{-3} \ M$.
Comparing the $[OH^-]$ concentrations: $2 \times 10^{-3} \ M > 10^{-3} \ M > \text{weak base concentrations}$.
Since $Ba(OH)_2$ provides the highest $[OH^-]$ concentration,it will have the highest $pOH$ and consequently the highest $pH$.

(C) The $pH$ scale ranges from $0$ to $14$.
Substances with $pH > 7$ are basic,and higher concentration of $OH^-$ ions leads to a higher $pH$.
$1 \, M \, NaOH$ is a strong base that completely dissociates to provide a high concentration of $OH^-$ ions,resulting in a $pH$ of $14$.
$1 \, M \, NH_3$ is a weak base with a $pH$ significantly lower than $14$.
Distilled water is neutral with a $pH$ of $7$.
Chlorine saturated water is acidic due to the formation of $HCl$ and $HOCl$,resulting in a $pH < 7$.
Therefore,$1 \, M \, NaOH$ has the highest $pH$.

(C) Initial concentration of $NaOH$ is $10^{-6} \ M$.
After diluting $100$ times,the new concentration is $[OH^-] = \frac{10^{-6}}{100} = 10^{-8} \ M$.
Since the concentration is very low,we must consider the contribution of $OH^-$ ions from water $(10^{-7} \ M)$.
Total $[OH^-] = 10^{-8} + 10^{-7} = 10^{-8} + 10 \times 10^{-8} = 11 \times 10^{-8} \ M$.
$pOH = -\log(11 \times 10^{-8}) = 8 - \log(11) \approx 8 - 1.04 = 6.96$.
$pH = 14 - pOH = 14 - 6.96 = 7.04$.
Therefore,the $pH$ lies between $7$ and $8$.

(A) Let the volume of each solution be $V \ L$.
For the first solution,$[H^+]_1 = 10^{-6} \ M$.
For the second solution,$[H^+]_2 = 10^{-3} \ M$.
When mixed,the total volume becomes $2V \ L$.
The total moles of $H^+$ ions = $(10^{-6} \times V) + (10^{-3} \times V) = V(10^{-6} + 10^{-3}) \ mol$.
The concentration of the resulting solution is $[H^+]_{mix} = \frac{V(10^{-6} + 10^{-3})}{2V} = \frac{10^{-3} + 0.001 \times 10^{-3}}{2} \approx \frac{10^{-3}}{2} = 0.5 \times 10^{-3} = 5 \times 10^{-4} \ M$.
Now,$pH = -\log[H^+]_{mix} = -\log(5 \times 10^{-4}) = -(\log 5 + \log 10^{-4}) = -(0.7 - 4) = 3.3$.

(A) For $HCl$,$pH = 1$ implies $[H^+] = 10^{-pH} = 10^{-1} = 0.1 \ M$.
Since $HCl$ is a monobasic acid,its normality $(N)$ is equal to its molarity $(M)$,so $N_{HCl} = 0.1 \ N$.
For $H_2SO_4$ to have the same $pH$,it must have the same concentration of $H^+$ ions,which is $[H^+] = 0.1 \ M$.
Normality is defined as the concentration of $H^+$ ions in the solution.
Therefore,the normality of the $H_2SO_4$ solution must be $0.1 \ N$.

(C) $1$. $H_2PO_4^-$ acts as an acid to form its conjugate base $HPO_4^{2-}$,so option $A$ is correct.
$2$. The ionic product of water $K_w = [H^+][OH^-] = 10^{-14}$ at $298 \ K$,hence $pH + pOH = 14$ is correct.
$3$. For a very dilute acid like $10^{-8} \ M \ HCl$,the contribution of $H^+$ from water cannot be ignored. The total $[H^+] = 10^{-8} + 10^{-7} \approx 1.1 \times 10^{-7} \ M$. Thus,$pH = -\log(1.1 \times 10^{-7}) \approx 6.96$,not $8$. Therefore,option $C$ is incorrect.
$4$. The auto-ionization of water is an endothermic process. As temperature increases,$K_w$ increases,which leads to a decrease in the $pH$ of neutral water. Thus,option $D$ is correct.

(D) $1$. Calculate the total moles of $H^+$ ions from $HCl$: $n(H^+) = M \times V \times \text{basicity} = 0.1 \times 0.010 \times 1 = 0.001 \ mol$.
$2$. Calculate the total moles of $H^+$ ions from $H_2SO_4$: $n(H^+) = M \times V \times \text{basicity} = 0.2 \times 0.040 \times 2 = 0.016 \ mol$.
$3$. Total moles of $H^+$ = $0.001 + 0.016 = 0.017 \ mol$.
$4$. Total volume of the mixture = $10 \ mL + 40 \ mL = 50 \ mL = 0.050 \ L$.
$5$. Concentration of $[H^+] = \frac{\text{Total moles}}{\text{Total volume}} = \frac{0.017}{0.050} = 0.34 \ M$.
$6$. $pH = -\log[H^+] = -\log(0.34) \approx 0.468 \approx 0.47$.

(D) The $pH$ of a solution is determined by the concentration of $H^+$ ions,which is defined as $pH = -\log[H^+]$.
Adding pure water (options $A$ and $B$) will dilute the solution,decreasing the concentration of $H^+$ ions and thus increasing the $pH$.
Adding concentrated $HCl$ (option $C$) will increase the concentration of $H^+$ ions,thus decreasing the $pH$.
Adding more of the same dilute $HCl$ (option $D$) does not change the concentration of $H^+$ ions in the solution,as the ratio of moles to volume remains constant. Therefore,the $pH$ remains unchanged.

(D) $H_2SO_4 \to 2H^{+} + SO_4^{2-}$
Since $H_2SO_4$ is a strong diprotic acid,it dissociates completely as $H_2SO_4 \to 2H^{+} + SO_4^{2-}$.
For a $1 \, M$ solution of $H_2SO_4$,the concentration of $H^{+}$ ions is $[H^{+}] = 2 \times 1 \, M = 2 \, M$.
The $pH$ is calculated as $pH = - \log [H^{+}]$.
$pH = - \log(2) = -0.3010$.

(C) $NaOH_{(aq)}$ is a strong base solution.
Since $NaOH$ dissociates completely,$[OH^-] = 0.01 \ M = 10^{-2} \ M$.
$pOH = -\log[OH^-] = -\log(10^{-2}) = 2$.
We know that $pH + pOH = 14$ at $25^{\circ}C$.
Therefore,$pH = 14 - pOH = 14 - 2 = 12$.

(D) For a solution to have $pH = 1$,the concentration of $[H^+]$ must be $10^{-1} \; M = 0.1 \; M$.
Let us calculate the final $[H^+]$ for each case:
$a. \; [H^+] = \frac{(60 \times 0.1) - (40 \times 0.1)}{60 + 40} = \frac{6 - 4}{100} = 0.02 \; M \; (pH \approx 1.7)$
$b. \; [H^+] = \frac{(55 \times 0.1) - (45 \times 0.1)}{55 + 45} = \frac{5.5 - 4.5}{100} = 0.01 \; M \; (pH = 2)$
$c. \; [H^+] = \frac{(75 \times 0.2) - (25 \times 0.2)}{75 + 25} = \frac{15 - 5}{100} = \frac{10}{100} = 0.1 \; M \; (pH = 1)$
$d. \; [H^+] = 0 \; M \; (pH = 7, \text{neutral solution})$
Thus,the solution $c$ has $pH = 1$.

For a very dilute acid solution,the contribution of $H_3O^+$ from the auto-ionization of water cannot be neglected.
$HCl_{(aq)} + H_2O_{(l)} \rightarrow H_3O^+_{(aq)} + Cl^-_{(aq)}$
$[H_3O^+]_{HCl} = 1.0 \times 10^{-8} \ M$
Let $x$ be the concentration of $H_3O^+$ produced by the auto-ionization of water: $2H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + OH^-_{(aq)}$.
Total $[H_3O^+] = (1.0 \times 10^{-8} + x)$ and $[OH^-] = x$.
$K_w = [H_3O^+][OH^-] = (1.0 \times 10^{-8} + x)(x) = 1.0 \times 10^{-14}$
$x^2 + 10^{-8}x - 10^{-14} = 0$
Using the quadratic formula $x = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{-10^{-8} + \sqrt{(10^{-8})^2 - 4(1)(-10^{-14})}}{2} = \frac{-10^{-8} + \sqrt{10^{-16} + 4 \times 10^{-14}}}{2} \approx 0.95 \times 10^{-7} \ M$
Total $[H_3O^+] = 1.0 \times 10^{-8} + 0.95 \times 10^{-7} = 1.05 \times 10^{-7} \ M$
$pH = -\log[H_3O^+] = -\log(1.05 \times 10^{-7}) \approx 6.98$.

$(i) \ 0.003 \ M \ HCl$
$HCl \rightarrow H^{+} + Cl^{-}$
$[H^{+}] = 0.003 \ M$
$pH = -\log[H^{+}] = -\log(3 \times 10^{-3}) = 3 - \log(3) = 3 - 0.477 = 2.523 \approx 2.52$
$(ii) \ 0.005 \ M \ NaOH$
$NaOH \rightarrow Na^{+} + OH^{-}$
$[OH^{-}] = 0.005 \ M$
$pOH = -\log[OH^{-}] = -\log(5 \times 10^{-3}) = 3 - \log(5) = 3 - 0.699 = 2.301$
$pH = 14 - pOH = 14 - 2.301 = 11.699 \approx 11.70$
$(iii) \ 0.002 \ M \ HBr$
$HBr \rightarrow H^{+} + Br^{-}$
$[H^{+}] = 0.002 \ M$
$pH = -\log[H^{+}] = -\log(2 \times 10^{-3}) = 3 - \log(2) = 3 - 0.301 = 2.699 \approx 2.70$
$(iv) \ 0.002 \ M \ KOH$
$KOH \rightarrow K^{+} + OH^{-}$
$[OH^{-}] = 0.002 \ M$
$pOH = -\log[OH^{-}] = -\log(2 \times 10^{-3}) = 3 - 0.301 = 2.699$
$pH = 14 - pOH = 14 - 2.699 = 11.301 \approx 11.30$

(N/A) Molar mass of $TlOH = 204.4 + 16 + 1 = 221.4 \,g/mol$. Concentration $[TlOH] = \frac{2 \,g}{221.4 \,g/mol \times 2 \,L} \approx 0.004516 \,M$. Since $TlOH$ is a strong base,$[OH^-] = 0.004516 \,M$. $pOH = -\log(0.004516) \approx 2.345$. $pH = 14 - 2.345 = 11.655$.
$(b)$ Molar mass of $Ca(OH)_2 = 40 + 2(16 + 1) = 74 \,g/mol$. Moles of $Ca(OH)_2 = \frac{0.3 \,g}{74 \,g/mol} \approx 0.004054 \,mol$. Concentration $[Ca(OH)_2] = \frac{0.004054 \,mol}{0.5 \,L} = 0.008108 \,M$. Since $Ca(OH)_2$ provides $2 \,OH^-$ ions,$[OH^-] = 2 \times 0.008108 = 0.016216 \,M$. $pOH = -\log(0.016216) \approx 1.79$. $pH = 14 - 1.79 = 12.21$.
$(c)$ Molar mass of $NaOH = 23 + 16 + 1 = 40 \,g/mol$. Moles of $NaOH = \frac{0.3 \,g}{40 \,g/mol} = 0.0075 \,mol$. Concentration $[NaOH] = \frac{0.0075 \,mol}{0.2 \,L} = 0.0375 \,M$. $[OH^-] = 0.0375 \,M$. $pOH = -\log(0.0375) \approx 1.426$. $pH = 14 - 1.426 = 12.574$.
$(d)$ Using dilution formula $M_1V_1 = M_2V_2$: $13.6 \,M \times 1 \,mL = M_2 \times 1000 \,mL$. $M_2 = 0.0136 \,M = [H^+]$. $pH = -\log(0.0136) \approx 1.866$.

(N/A) $1$. Calculate the molarity of $KOH$ solution:
$M = \frac{\text{mass in } g}{\text{molar mass} \times \text{volume in } L} = \frac{0.561 \, g}{56.11 \, g/mol \times 0.200 \, L} = 0.05 \, M$
$2$. Since $KOH$ is a strong base,it dissociates completely:
$KOH_{(aq)} \to K^{+}_{(aq)} + OH^{-}_{(aq)}$
Therefore,$[K^{+}] = 0.05 \, M$ and $[OH^{-}] = 0.05 \, M$.
$3$. Calculate $[H^{+}]$ using the ionic product of water ($K_w = 10^{-14}$ at $298 \, K$):
$[H^{+}] = \frac{K_w}{[OH^{-}]} = \frac{10^{-14}}{0.05} = 2 \times 10^{-13} \, M$
$4$. Calculate $pH$:
$pH = -\log[H^{+}] = -\log(2 \times 10^{-13}) = 13 - \log(2) = 13 - 0.3010 = 12.699 \approx 12.70$

(N/A) Molar mass of $Sr(OH)_2 = 87.62 + 2 \times (16.00 + 1.01) = 121.64 \, g/mol$.
Concentration of $Sr(OH)_2 = \frac{19.23 \, g/L}{121.64 \, g/mol} = 0.1581 \, M$.
Dissociation reaction: $Sr(OH)_{2(aq)} \to Sr^{2+}_{(aq)} + 2OH^{-}_{(aq)}$.
Concentration of strontium ions: $[Sr^{2+}] = 0.1581 \, M$.
Concentration of hydroxyl ions: $[OH^{-}] = 2 \times 0.1581 \, M = 0.3162 \, M$.
To find $pH$: $pOH = -\log[OH^{-}] = -\log(0.3162) \approx 0.50$.
$pH = 14 - pOH = 14 - 0.50 = 13.50$.

(B) $NaOH$ is a strong base and dissociates completely as: $NaOH \rightarrow Na^+ + OH^-$.
Concentration of $[OH^-] = 0.03 \ M$.
$pOH = -\log[OH^-] = -\log(0.03) = -(\log 3 + \log 10^{-2}) = -(0.477 - 2) = 1.523 \approx 1.52$.
Since $pH + pOH = 14$ at $298 \ K$,
$pH = 14 - 1.52 = 12.48$.
Therefore,the $pH$ is $12.48$ and the $pOH$ is $1.52$.

(A) Step $1$: Calculate the initial moles of $H^+$. $n = M \times V = 0.001 \ mol/L \times 0.0001 \ L = 10^{-7} \ mol$.
Step $2$: Calculate the final concentration of $H^+$ in $10 \ L$ solution. $[H^+] = \frac{10^{-7} \ mol}{10 \ L} = 10^{-8} \ M$.
Step $3$: Since the concentration of $H^+$ from $HCl$ is very low $(10^{-8} \ M)$,we must consider the contribution of $H^+$ from water $(10^{-7} \ M)$.
Total $[H^+] = 10^{-8} + 10^{-7} = 1.1 \times 10^{-7} \ M$.
Step $4$: Calculate $pH = -\log[H^+] = -\log(1.1 \times 10^{-7}) = 7 - \log(1.1) \approx 7 - 0.04 = 6.96$.

(A) For $HCl$ (strong acid),$[H^+] = 0.1 \ M = 10^{-1} \ M$. $pH = -\log[H^+] = -\log(10^{-1}) = 1$.
$(b)$ For $H_2SO_4$ (strong acid),$[H^+] = 2 \times 0.1 \ M = 0.2 \ M$. $pH = -\log(0.2) = -(\log 2 - 1) = 1 - 0.3010 = 0.6990$.
$(c)$ For $HNO_3$ (strong acid),$[H^+] = 0.1 \ M = 10^{-1} \ M$. $pH = -\log(10^{-1}) = 1$.
$(d)$ For $NaOH$ (strong base),$[OH^-] = 0.1 \ M = 10^{-1} \ M$. $pOH = -\log(10^{-1}) = 1$. $pH = 14 - pOH = 14 - 1 = 13$.
$(e)$ For $KOH$ (strong base),$[OH^-] = 0.1 \ M = 10^{-1} \ M$. $pOH = -\log(10^{-1}) = 1$. $pH = 14 - pOH = 14 - 1 = 13$.
$(f)$ For $Ba(OH)_2$ (strong base),$[OH^-] = 2 \times 0.1 \ M = 0.2 \ M$. $pOH = -\log(0.2) = 0.6990$. $pH = 14 - 0.6990 = 13.3010$.

(N/A) Initial state: $V_1 = 1.0 \ mL$,$M_1 = 0.1 \ M$
Final state: $V_2 = 50 \ mL$,$M_2 = ?$
Using the dilution equation: $M_1 V_1 = M_2 V_2$
$0.1 \ M \times 1.0 \ mL = M_2 \times 50 \ mL$
$M_2 = \frac{0.1}{50} = 0.002 \ M = 2 \times 10^{-3} \ M$
Initial $pH = -\log(0.1) = 1$
Final $pH = -\log(2 \times 10^{-3}) = 3 - \log(2) = 3 - 0.301 = 2.699$
Change in $pH = 2.699 - 1 = 1.699$