(B) For a weak acid $HA$ titrated with $NaOH$,the $pH$ is given by the Henderson-Hasselbalch equation: $pH = pK_{a} + \log \frac{[Salt]}{[Acid]}$.Give

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(B) For a weak acid $HA$ titrated with $NaOH$,the $pH$ is given by the Henderson-Hasselbalch equation: $pH = pK_{a} + \log \frac{[Salt]}{[Acid]}$.Given $K_{a} = 10^{-5}$,so $pK_{a} = -\log(10^{-5}) = 5$.At one-third neutralization,the amount of salt formed is $1/3$ and the remaining acid is $1 - 1/3 = 2/3$.Substituting these values into the equation: $pH = 5 + \log \frac{1/3}{2/3} = 5 + \log(1/2) = 5 - \log 2$.

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)Buffer solution

Difficult

$A$ weak acid of dissociation constant $10^{-5}$ is being titrated with aqueous $NaOH$ solution. The $pH$ at the point of one-third neutralization of the acid will be:-

A
$5 + \log 2 - \log 3$
B
$5 - \log 2$
C
$5 - \log 3$
D
$5 - \log 6$

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6-2.Equilibrium-II (Ionic Equilibrium)

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Buffer solution

Calculate the amount of $(NH_4)_2SO_4$ in grams which must be added to $500 \ mL$ of $0.200 \ M \ NH_3$ to yield a solution with $pH = 9.35$ ($K_b$ for $NH_3 = 1.78 \times 10^{-5}$).

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If $50 \ mL$ of $0.2 \ M \ KOH$ is added to $40 \ mL$ of $0.5 \ M \ HCOOH,$ the $pH$ of the resulting solution is $(K_a = 1.8 \times 10^{-4})$

Medium

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$A$ buffer solution is prepared by mixing $0.2 \ M \ NH_4OH$ and $1 \ M \ NH_4Cl$. What is the $pH$ value of the buffer solution? (Given $pK_b = 4.744$)

DifficultMHT CET 2024

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$0.1 \ mol$ of $CH_3NH_2$ $(K_b = 5 \times 10^{-4})$ is mixed with $0.08 \ mol$ of $HCl$ and diluted to $1 \ L$. What will be the $[H^{+}]$ concentration in the solution?

DifficultIIT 2005

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The pH of a solution obtained by mixing $5 \text{ mL}$ of $0.1 \text{ M } NH_4OH$ solution with $250 \text{ mL}$ of $0.1 \text{ M } NH_4Cl$ solution is . . . . . . $\times 10^{-2}$.

DifficultJEE MAIN 2026

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$A$ weak acid of dissociation constant $10^{-5}$ is being titrated with aqueous $NaOH$ solution. The $pH$ at the point of one-third neutralization of the acid will be:-

Similar Questions

Calculate the amount of $(NH_4)_2SO_4$ in grams which must be added to $500 \ mL$ of $0.200 \ M \ NH_3$ to yield a solution with $pH = 9.35$ ($K_b$ for $NH_3 = 1.78 \times 10^{-5}$).

If $50 \ mL$ of $0.2 \ M \ KOH$ is added to $40 \ mL$ of $0.5 \ M \ HCOOH,$ the $pH$ of the resulting solution is $(K_a = 1.8 \times 10^{-4})$

$A$ buffer solution is prepared by mixing $0.2 \ M \ NH_4OH$ and $1 \ M \ NH_4Cl$. What is the $pH$ value of the buffer solution? (Given $pK_b = 4.744$)

$0.1 \ mol$ of $CH_3NH_2$ $(K_b = 5 \times 10^{-4})$ is mixed with $0.08 \ mol$ of $HCl$ and diluted to $1 \ L$. What will be the $[H^{+}]$ concentration in the solution?

The pH of a solution obtained by mixing $5 \text{ mL}$ of $0.1 \text{ M } NH_4OH$ solution with $250 \text{ mL}$ of $0.1 \text{ M } NH_4Cl$ solution is . . . . . . $\times 10^{-2}$.

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