The Henderson-Hasselbalch equation for an aci | 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: $pH = pK_a + \log \frac{[	ext{Conjugate Base}]}{[	ext{Acid}]}$ Rearranging this,we g

Vedclass · Accepted Answer

$[	ext{Acid}] 	imes 10^{-5} = [	ext{Conjugate Base}]$

Vedclass · Answer

(B) The Henderson-Hasselbalch equation for an acidic buffer is: $pH = pK_a + \log \frac{[	ext{Conjugate Base}]}{[	ext{Acid}]}$ Rearranging this,we get: $pH - pK_a = \log \frac{[	ext{Conjugate Base}]}{[	ext{Acid}]}$ Given that $pH - pK_a = 5$,we substitute this into the equation: $5 = \log \frac{[	ext{Conjugate Base}]}{[	ext{Acid}]}$ Taking the antilog on both sides: $10^5 = \frac{[	ext{Conjugate Base}]}{[	ext{Acid}]}$ Therefore,$[	ext{Conjugate Base}] = [	ext{Acid}] 	imes 10^5$ Or,$[	ext{Acid}] = [	ext{Conjugate Base}] 	imes 10^{-5}$ Comparing this with the given options,option $B$ is correct.

The Henderson-Hasselbalch equation for an acidic buffer is given by $pH - pK_a = 5$. This condition holds true when:

Similar Questions

$A$ buffer solution contains equal concentrations of weak acid and its salt with a strong base. Calculate the $pH$ of the buffer solution if the dissociation constant of the weak acid is $1.8 \times 10^{-5}$.

If $12.5 \ mL$ of $0.1 \ M$ $KOH$ is added into $50 \ mL$ of $0.1 \ M$ $CH_3COOH$ solution,then what would be the final $pH$? $(K_a(CH_3COOH) = 2 \times 10^{-5})$

When $500 \, mL$ of $0.2 \, M$ acetic acid is added to $500 \, mL$ of $0.30 \, M$ sodium acetate solution,what is the $pH$ of the resulting solution? (Given: Dissociation constant of acetic acid $K_a = 1.5 \times 10^{-5}$)

$20 \ mL$ of $0.1 \ M$ acetic acid is mixed with $50 \ mL$ of potassium acetate. $K_a$ of acetic acid $= 1.8 \times 10^{-5}$ at $27^{\circ} C$. Calculate the concentration of potassium acetate if the $pH$ of the mixture is $4.8$. (in $M$)

Which among the following pairs is not an acidic buffer?

Vedclass Test Series

Exam Paper Generator

Online Exam Module