In a mixture of a weak acid and its salt,the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For a buffer solution containing a weak acid and its salt,the Henderson-Hasselbalch equation is given by: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$

Vedclass · Accepted Answer

Decreases by $1$

Vedclass · Answer

(A) For a buffer solution containing a weak acid and its salt,the Henderson-Hasselbalch equation is given by: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.Let the initial ratio be $R_1 = \frac{[Salt]}{[Acid]}$. Then $pH_1 = pK_a + \log(R_1)$.The problem states that the ratio of acid to salt is increased ten-fold,which means the new ratio $\frac{[Acid]}{[Salt]}$ is $10 	imes \frac{[Acid]_{initial}}{[Salt]_{initial}}$.Therefore,the new ratio of salt to acid is $R_2 = \frac{[Salt]}{[Acid]} = \frac{1}{10} 	imes R_1$.The new $pH$ is $pH_2 = pK_a + \log(R_2) = pK_a + \log(\frac{R_1}{10}) = pK_a + \log(R_1) - \log(10)$.Since $\log(10) = 1$,we get $pH_2 = pH_1 - 1$.Thus,the $pH$ decreases by $1$.

In a mixture of a weak acid and its salt,the ratio of concentration of acid to salt is increased ten-fold. The $pH$ of the solution

Similar Questions

In which of the following does a buffer solution play an important role?

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 $NH_3 = 1.78 \times 10^{-5}$).

Which of the following is not a buffer solution?

$A$ buffer solution has equal volumes of $0.2 \ M \ NH_4OH$ and $0.02 \ M \ NH_4Cl$. The $pK_b$ of the base is $5$. The $pH$ is

The ratio of volumes of $CH_3COOH$ $0.1 \ N$ to $CH_3COONa$ $0.1 \ N$ required to prepare a buffer solution of $pH$ $5.74$ is (Given, $pK_a$ of $CH_3COOH$ is $4.74$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module