To prepare a buffer solution of pH = 6 by mix | vedclass

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

Question

(B) For an acidic buffer,the Henderson-Hasselbalch equation is given by: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$Given: $pH = 6$ and $K_a = 10^{-5}$.F

Vedclass · Accepted Answer

$10 : 1$

Vedclass · Answer

(B) For an acidic buffer,the Henderson-Hasselbalch equation is given by: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$Given: $pH = 6$ and $K_a = 10^{-5}$.First,calculate $pK_a$: $pK_a = -\log(K_a) = -\log(10^{-5}) = 5$.Substitute the values into the equation: $6 = 5 + \log \frac{[Salt]}{[Acid]}$$6 - 5 = \log \frac{[Salt]}{[Acid]}$$1 = \log \frac{[Salt]}{[Acid]}$Taking the antilog on both sides: $\frac{[Salt]}{[Acid]} = 10^1 = 10$.Therefore,the ratio of salt to acid is $10 : 1$.

To prepare a buffer solution of $pH = 6$ by mixing sodium acetate and acetic acid,what should be the ratio of the concentration of salt to acid? $(K_a = 10^{-5})$

Similar Questions

The acidic and basic strength of buffer solutions remains constant because ...

Which of the following mixtures act$(s)$ as a buffer solution?

How to prepare a buffer solution using ammonia $(pK_b = 4.75)$? What is the $pH$ of this buffer if the concentrations of ammonia and ammonium chloride are equal?

$A$ weak acid,$HA$,is found to be $10\%$ ionized in $0.01\ M$ aqueous solution. Calculate the $pH$ of a solution which is $0.1\ M$ in $HA$ and $0.05\ M$ in $NaA$.

The $pK_b$ of $CN^{\Theta}$ is $4.7$. The $pH$ of a solution prepared by mixing $2.5 \ mol$ of $KCN$ and $2.5 \ mol$ of $HCN$ in water and making the total volume up to $500 \ mL$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module