A student needs to prepare a buffer solution | vedclass

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

Question

(B) The Henderson-Hasselbalch equation for an acidic buffer is given by:$pH = pK_{a} + \log \frac{[Salt]}{[Acid]}$First,calculate $pK_{a}$:$pK_{a} = -

Vedclass · Accepted Answer

$0.13$

Vedclass · Answer

(B) The Henderson-Hasselbalch equation for an acidic buffer is given by:$pH = pK_{a} + \log \frac{[Salt]}{[Acid]}$First,calculate $pK_{a}$:$pK_{a} = -\log(K_{a}) = -\log(1.3 	imes 10^{-5}) = 5 - \log(1.3) \approx 5 - 0.1139 = 4.8861$Substitute the values into the equation:$4 = 4.8861 + \log \frac{[CH_{3}CH_{2}COO^{-}]}{[CH_{3}CH_{2}COOH]}$$\log \frac{[CH_{3}CH_{2}COO^{-}]}{[CH_{3}CH_{2}COOH]} = 4 - 4.8861 = -0.8861$Taking the antilog:$\frac{[CH_{3}CH_{2}COO^{-}]}{[CH_{3}CH_{2}COOH]} = 10^{-0.8861} \approx 0.13$

$A$ student needs to prepare a buffer solution of propanoic acid and its sodium salt with $pH$ $4$. The ratio of $\frac{[CH_{3}CH_{2}COO^{-}]}{[CH_{3}CH_{2}COOH]}$ required to make the buffer is .....
Given: $K_{a}(CH_{3}CH_{2}COOH) = 1.3 \times 10^{-5}$

Similar Questions

Which of the following mixtures forms an acidic buffer?

The dissociation constant of $HCN$ is $5 \times 10^{-10}$. The $pH$ of the solution prepared by mixing $1.5 \, mole$ of $HCN$ and $0.15 \, mole$ of $KCN$ in water and making up the total volume to $0.5 \, dm^3$ is (in $.302$)

The $pKa$ of a weak acid $HA$ is $4.5$. The $pOH$ of an aqueous buffer solution in which the acid $HA$ is $50\%$ ionized is:

Calculate $pH$ of the solution obtained by mixing $50 \ mL$ of $0.2 \ M$ $NH_4Cl$ solution and $75 \ mL$ of $0.1 \ M$ $NaOH$ solution. Given $pK_b$ for aqueous $NH_3$ is $4.74$.

Henderson's equation is $pH = pK_a + \log\frac{[\text{salt}]}{[\text{acid}]}$. If the acid is half-neutralized,the value of $pH$ will be: $[pK_a = 4.30]$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$A$ student needs to prepare a buffer solution of propanoic acid and its sodium salt with $pH$ $4$. The ratio of $\frac{[CH_{3}CH_{2}COO^{-}]}{[CH_{3}CH_{2}COOH]}$ required to make the buffer is .....Given: $K_{a}(CH_{3}CH_{2}COOH) = 1.3 \times 10^{-5}$