pH of strong Acids and strong Bases Questions in English

(B) For solution $I, pH=3$
$\Rightarrow [H^{+}] = 10^{-3} \ M$
$V_{1} = 100 \ mL$
For solution $II, pH = 4$
$\Rightarrow [H^{+}] = 10^{-4} \ M$
$V_{2} = 400 \ mL$
Concentration of resulting solution $M = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}$
$M = \frac{10^{-3} \times 100 + 10^{-4} \times 400}{100 + 400}$
$M = \frac{0.1 + 0.04}{500} = \frac{0.14}{500} = 2.8 \times 10^{-4} \ M$
$[H^{+}] = 2.8 \times 10^{-4} \ M$
$pH = -\log [H^{+}] = -\log (2.8 \times 10^{-4})$
$pH = 4 - \log 2.8$

(A) For a solution to have $pH = 1$,the concentration of $[H^+]$ must be $0.1 \ M$.
$(A)$ $0.1 \ M \ CH_3COOH$ is a weak acid and does not dissociate completely,so $[H^+] < 0.1 \ M$,hence $pH > 1$.
$(B)$ $0.1 \ M \ HNO_3$ is a strong acid,$[H^+] = 0.1 \ M$,so $pH = -\log(0.1) = 1$.
$(C)$ $0.05 \ M \ H_2SO_4$ is a strong acid,$[H^+] = 2 \times 0.05 = 0.1 \ M$,so $pH = -\log(0.1) = 1$.
$(D)$ For the mixture: $n(H^+) = 50 \times 0.4 = 20 \ mmol$,$n(OH^-) = 50 \times 0.2 = 10 \ mmol$. Remaining $n(H^+) = 20 - 10 = 10 \ mmol$. Total volume = $100 \ cm^3$. $[H^+] = 10 / 100 = 0.1 \ M$,so $pH = 1$.

(D) The chemical reaction is: $2Na + 2H_{2}O \longrightarrow 2NaOH + H_{2}$
Molar mass of $Na = 23 \ g/mol$.
Moles of $Na = \frac{0.023 \ g}{23 \ g/mol} = 0.001 \ mol$.
According to the stoichiometry,$2 \ mol$ of $Na$ produces $2 \ mol$ of $NaOH$.
Therefore,$0.001 \ mol$ of $Na$ produces $0.001 \ mol$ of $NaOH$.
The volume of the solution is $100 \ cm^{3} = 0.1 \ L$.
Concentration of $[OH^{-}] = \frac{0.001 \ mol}{0.1 \ L} = 0.01 \ M = 10^{-2} \ M$.
$pOH = -\log[OH^{-}] = -\log(10^{-2}) = 2$.
Since $pH + pOH = 14$,then $pH = 14 - 2 = 12$.

(D) The chemical reaction is: $2Na + 2H_2O \rightarrow 2NaOH + H_2$
Number of moles of $Na = \frac{0.023 \ g}{23 \ g/mol} = 1 \times 10^{-3} \ mol$.
According to the stoichiometry,$2 \ mol$ of $Na$ produces $2 \ mol$ of $NaOH$.
Therefore,$1 \times 10^{-3} \ mol$ of $Na$ produces $1 \times 10^{-3} \ mol$ of $NaOH$.
The volume of the solution is $100 \ cm^3 = 0.1 \ L$.
Concentration of $[OH^-] = \frac{1 \times 10^{-3} \ mol}{0.1 \ L} = 1 \times 10^{-2} \ M$.
$pOH = -\log[OH^-] = -\log(1 \times 10^{-2}) = 2$.
Since $pH + pOH = 14$,$pH = 14 - 2 = 12$.

(A) Initial concentration of $[OH^-] = 10^{-6} \ M$.
Upon dilution by $100$ times,the concentration becomes $[OH^-] = \frac{10^{-6}}{100} = 10^{-8} \ M$.
Since this concentration is very low,we must consider the contribution of $[OH^-]$ from the auto-ionization of water,which is $10^{-7} \ M$.
Total $[OH^-] = 10^{-8} + 10^{-7} = 10^{-8} (1 + 10) = 11 \times 10^{-8} \ M$.
$pOH = -\log(11 \times 10^{-8}) = 8 - \log(11) \approx 8 - 1.0414 = 6.9586$.
$pH = 14 - pOH = 14 - 6.9586 = 7.0414$.
Thus,the $pH$ lies between $7$ and $8$.

(C) Lime water,$Ca(OH)_2$,is a strong base solution.
According to the law of equivalence,the normality $(N)$ of a strong base is equal to the concentration of hydroxide ions $[OH^-]$.
$[OH^-] = N = 0.01 \ N = 10^{-2} \ M$.
$pOH = -\log [OH^-] = -\log(10^{-2}) = 2$.
Since $pH + pOH = 14$ at $25^{\circ}C$,
$pH = 14 - pOH = 14 - 2 = 12$.

(A) Moles of $H^{+}$ from $HCl = 1 \ L \times 0.02 \ M = 0.02 \ mol$.
Moles of $H^{+}$ from $H_2SO_4 = 1 \ L \times 0.01 \ M \times 2 = 0.02 \ mol$.
Total moles of $H^{+} = 0.02 + 0.02 = 0.04 \ mol$.
Total volume of the mixture $= 1 \ L + 1 \ L = 2 \ L$.
Concentration of $[H^{+}] = \frac{0.04 \ mol}{2 \ L} = 0.02 \ M$.
$pH = -\log_{10}[H^{+}] = -\log_{10}(0.02) = -\log_{10}(2 \times 10^{-2})$.
$pH = -(\log_{10} 2 + \log_{10} 10^{-2}) = -(0.3 - 2) = 1.7$.

(A) For a strong acid like $HCl$,it dissociates completely in water: $HCl \rightarrow H^{+} + Cl^{-}$.
Given concentration of $HCl = 0.10 \ M$,so $[H^{+}] = 0.10 \ M = 10^{-1} \ M$.
The $pH$ of the solution is calculated as: $pH = -\log[H^{+}] = -\log(10^{-1}) = 1$.
Using the relation at $25^{\circ}C$: $pH + pOH = 14$.
Therefore,$pOH = 14 - pH = 14 - 1 = 13$.

(C) The $pH$ scale ranges from $0$ to $14$. Substances with $pH > 7$ are basic,and higher $pH$ values indicate stronger basicity.
$1 \ M$ $HCl$ is a strong acid $(pH \approx 0)$.
Distilled $H_2O$ is neutral $(pH = 7)$.
$1 \ M$ $aq$ $NH_3$ is a weak base $(pH \approx 11.6)$.
$1 \ M$ $NaOH$ is a strong base $(pH = 14)$.
Therefore,$1 \ M$ $NaOH$ has the highest $pH$.

(D) $HCl$ is a strong acid. All acids have a $pH$ in the acidic range,i.e.,below $7$.
The concentration of $HCl$ is given as $10^{-8} \ M$.
$pH$ is calculated by considering the contribution of protons from both the acid and the water.
Since $10^{-8} \ M$ is a very low concentration,the contribution of $H^+$ ions from the auto-ionization of water $(10^{-7} \ M)$ cannot be ignored.
Total $[H^+] = [H^+]_{acid} + [H^+]_{water} = 10^{-8} + 10^{-7} = 1.1 \times 10^{-7} \ M$.
$pH = -\log[H^+] = -\log(1.1 \times 10^{-7}) = 7 - \log(1.1) \approx 6.96$.
This value lies between $6$ and $7$.
Thus,the correct answer is option $(d)$.

(D) The molecular mass of $NaOH = 23 + 16 + 1 = 40 \ g/mol$.
The number of moles of $NaOH = \frac{0.04 \ g}{40 \ g/mol} = 0.001 \ mol = 10^{-3} \ mol$.
The volume of the solution = $100 \ mL = 0.1 \ L$.
Molarity of $NaOH$ solution = $\frac{10^{-3} \ mol}{0.1 \ L} = 10^{-2} \ M$.
Since $NaOH$ is a strong base,$[OH^-] = 10^{-2} \ M$.
$pOH = -\log[OH^-] = -\log(10^{-2}) = 2$.
Using the relation $pH + pOH = 14$,we get $pH = 14 - 2 = 12$.

(C) $H_2SO_4$ is a strong diprotic acid,so it dissociates completely as: $H_2SO_4 \rightarrow 2H^{+} + SO_4^{2-}$.
Concentration of $[H^{+}] = 2 \times 0.001 \ M = 0.002 \ M = 2 \times 10^{-3} \ M$.
Using the ionic product of water at $25^{\circ}C$,$K_w = [H^{+}][OH^{-}] = 10^{-14}$.
$[OH^{-}] = \frac{K_w}{[H^{+}]} = \frac{10^{-14}}{2 \times 10^{-3}}$.
$[OH^{-}] = 0.5 \times 10^{-11} \ M = 5 \times 10^{-12} \ M$.

(D) Since $HCl$ is a strong acid,it dissociates completely in water.
However,for very dilute solutions $(10^{-8} \ M)$,the contribution of $H^+$ ions from the auto-ionization of water cannot be neglected.
Let the concentration of $H^+$ ions from water be $x$.
The total concentration of $H^+$ ions is $[H^+] = 10^{-8} + x$.
The concentration of $OH^-$ ions from water is $[OH^-] = x$.
Using the ionic product of water,$K_w = [H^+][OH^-] = 10^{-14}$.
$(10^{-8} + x)(x) = 10^{-14} \implies x^2 + 10^{-8}x - 10^{-14} = 0$.
Solving this quadratic equation for $x$ using the quadratic formula $x = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{-10^{-8} + \sqrt{(10^{-8})^2 - 4(1)(-10^{-14})}}{2} = \frac{-10^{-8} + \sqrt{10^{-16} + 4 \times 10^{-14}}}{2} \approx 0.95 \times 10^{-8} \ M$.
Total $[H^+] = 10^{-8} + 0.95 \times 10^{-8} = 1.05 \times 10^{-8} \ M$ (Correction: $1.05 \times 10^{-7} \ M$ is the correct total concentration).
$pH = -\log[H^+] = -\log(1.05 \times 10^{-7}) \approx 6.98 \approx 6.9$.
Thus,option $(D)$ is the correct answer.

(C) Barium hydroxide is a strong base and dissociates completely as: $Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^-$.
Initial concentration of $Ba(OH)_2 = 1 \times 10^{-3} \ M$.
Initial concentration of $OH^- = 2 \times 1 \times 10^{-3} = 2 \times 10^{-3} \ M$.
Upon adding $50 \ mL$ of $H_2O$ to $50 \ mL$ of the solution,the total volume becomes $100 \ mL$.
The new concentration of $OH^-$ is calculated using the dilution formula $M_1V_1 = M_2V_2$:
$(2 \times 10^{-3} \ M) \times (50 \ mL) = M_2 \times (100 \ mL)$.
$M_2 = \frac{2 \times 10^{-3} \times 50}{100} = 1 \times 10^{-3} \ M$.
$pOH = -\log[OH^-] = -\log(1 \times 10^{-3}) = 3$.
$pH = 14 - pOH = 14 - 3 = 11.0$.

(A) Step $1$: Calculate the total moles of $HCl$ in the mixture.
$n_1 = M_1 \times V_1 = 0.2 \ M \times 75 \ mL = 15 \ mmol$
$n_2 = M_2 \times V_2 = 1 \ M \times 25 \ mL = 25 \ mmol$
Total moles of $HCl = n_1 + n_2 = 15 + 25 = 40 \ mmol$.
Step $2$: Calculate the total volume of the final solution.
$V_{total} = V_1 + V_2 + V_{water} = 75 \ mL + 25 \ mL + 300 \ mL = 400 \ mL$.
Step $3$: Calculate the final molarity $(M_{final})$ of $HCl$.
$M_{final} = \frac{\text{Total moles}}{\text{Total volume}} = \frac{40 \ mmol}{400 \ mL} = 0.1 \ M$.
Step $4$: Calculate the $pH$ of the solution.
Since $HCl$ is a strong acid,$[H^+] = [HCl] = 0.1 \ M = 10^{-1} \ M$.
$pH = -\log_{10}[H^+] = -\log_{10}(10^{-1}) = 1$.

(B) For $HCl$ solution: $pH = 2$,so $[H^+] = 10^{-2} \ M$. Moles of $H^+ = 10^{-2} \ mol/L \times 0.2 \ L = 2 \times 10^{-3} \ mol$.
For $NaOH$ solution: $pH = 12$,so $pOH = 14 - 12 = 2$. $[OH^-] = 10^{-2} \ M$. Moles of $OH^- = 10^{-2} \ mol/L \times 0.3 \ L = 3 \times 10^{-3} \ mol$.
Since $n(OH^-) > n(H^+)$,the reaction $H^+ + OH^- \rightarrow H_2O$ occurs.
Remaining moles of $OH^- = 3 \times 10^{-3} - 2 \times 10^{-3} = 1 \times 10^{-3} \ mol$.
The total volume of the resulting solution is $1.0 \ L$.
Concentration of $OH^- = \frac{1 \times 10^{-3} \ mol}{1.0 \ L} = 10^{-3} \ M$.
$pOH = -\log(10^{-3}) = 3$.
$pH = 14 - pOH = 14 - 3 = 11$.

(B) Initial moles of $HCl$ $(n_1)$ = $0.5 \ M \times 0.1 \ L = 0.05 \ mol$.
Initial moles of $NaOH$ $(n_2)$ = $0.4 \ M \times 0.1 \ L = 0.04 \ mol$.
Since $HCl$ is a strong acid and $NaOH$ is a strong base,they neutralize each other: $n_{H^+} = n_1 - n_2 = 0.05 - 0.04 = 0.01 \ mol$.
Total volume of the final solution = $100 \ mL + 100 \ mL + 800 \ mL = 1000 \ mL = 1 \ L$.
Final concentration of $H^+$ ions = $\frac{0.01 \ mol}{1 \ L} = 0.01 \ M = 10^{-2} \ M$.
$pH = -\log[H^+] = -\log(10^{-2}) = 2$.

(C) For solution $A$: $pH=6.0$,so $[H^{+}]_A = 10^{-6} \ M$.
For solution $B$: $pH=4.0$,so $[H^{+}]_B = 10^{-4} \ M$.
Let the volume of each solution be $1 \ L$.
Total moles of $H^{+}$ = $(1 \ L \times 10^{-6} \ M) + (1 \ L \times 10^{-4} \ M) = 10^{-6} + 10^{-4} = 1.01 \times 10^{-4} \ \text{moles}$.
Total volume of the mixture = $1 \ L + 1 \ L = 2 \ L$.
Resultant $[H^{+}] = \frac{1.01 \times 10^{-4} \ \text{moles}}{2 \ L} = 5.05 \times 10^{-5} \ M$.
$pH = -\log(5.05 \times 10^{-5}) = 5 - \log(5.05) \approx 5 - 0.703 = 4.297$.
Since $4.297$ lies between $4$ and $5$,the correct range is between $4$ and $5$.

(D) For a very dilute solution like $10^{-8} \ M \ HCl$,the concentration of $H^{+}$ ions from water cannot be neglected.
$H^{+}_{total} = [H^{+}]_{HCl} + [H^{+}]_{H_2O}$
$[H^{+}]_{total} = 10^{-8} + 10^{-7} \approx 1.1 \times 10^{-7} \ M$
$pH = -\log[H^{+}]_{total} = -\log(1.1 \times 10^{-7})$
$pH = 7 - \log(1.1) \approx 7 - 0.0414 = 6.9586$
Thus,the pH is greater than $6$ and less than $7$.

(D) $KOH$ is a strong base,so $[OH^-] = [KOH] = 10^{-4} \ M$.
$pOH = -\log[OH^-] = -\log(10^{-4}) = 4$.
At $25^\circ C$,$pH + pOH = 14$.
$pH = 14 - 4 = 10$.

(B) For a very dilute solution of a strong base,the contribution of $OH^{-}$ ions from the dissociation of water cannot be neglected.
$KOH \longrightarrow K^{+} + OH^{-}$
$[OH^{-}]_{KOH} = 10^{-8} \ M$
$[OH^{-}]_{water} = 10^{-7} \ M$
Total $[OH^{-}] = 10^{-8} + 10^{-7} = 10^{-7}(0.1 + 1) = 1.1 \times 10^{-7} \ M$
$pOH = -\log(1.1 \times 10^{-7}) = 7 - \log(1.1) \approx 7 - 0.0414 = 6.9586$
Since $pH + pOH = 14$ at $25^{\circ} C$,
$pH = 14 - 6.9586 = 7.0414 \approx 7.04$.
Given the options,the closest value is $7.02$.

(B) Step $1$: Calculate the total millimoles of $H^{+}$ ions.
Millimoles of $H^{+}$ from $HCl = 0.1 \ (M) \times 2 \ mL \times 1 = 0.2 \ mmol$.
Millimoles of $H^{+}$ from $H_2SO_4 = 0.1 \ (M) \times 2 \ mL \times 2 = 0.4 \ mmol$.
Total millimoles of $H^{+} = 0.2 + 0.4 = 0.6 \ mmol$.
Step $2$: Calculate the concentration of $H^{+}$ in the final mixture.
Total volume $= 2 \ mL + 2 \ mL + 2 \ mL = 6 \ mL$.
$[H^{+}] = \frac{\text{Total millimoles}}{\text{Total volume in } mL} = \frac{0.6}{6} = 0.1 \ (M)$.
Step $3$: Calculate the $pH$.
$pH = -\log_{10} [H^{+}] = -\log_{10} (0.1) = 1$.

(A) The initial $pH$ of $0.01 \ M \ HCl$ is $-\log(0.01) = 2$.
To decrease the $pH$,the concentration of $H^+$ ions must increase.
Option $A$: Adding $5 \ mL$ of $1 \ M \ HCl$ adds $5 \ mmol$ of $H^+$ to the existing $0.5 \ mmol$. The new concentration becomes $\frac{5.5 \ mmol}{55 \ mL} = 0.1 \ M$. The new $pH$ is $-\log(0.1) = 1$. Since $1 < 2$,the $pH$ decreases.
Option $B$: Adding $50 \ mL$ of $0.01 \ M \ HCl$ results in the same concentration $(0.01 \ M)$,so the $pH$ remains $2$.
Option $C$: Adding $50 \ mL$ of $0.002 \ M \ HCl$ dilutes the solution,decreasing the $H^+$ concentration and increasing the $pH$.
Option $D$: Adding $Mg$ reacts with $HCl$ to consume $H^+$ ions,increasing the $pH$.