For a buffer solution containing equal concen | vedclass

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

Question

(A) Given that the concentration of the base $B^-$ and its conjugate acid $HB$ are equal,i.e.,$[B^-] = [HB]$.
For a basic buffer,the $pOH$ is given b

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) Given that the concentration of the base $B^-$ and its conjugate acid $HB$ are equal,i.e.,$[B^-] = [HB]$.
For a basic buffer,the $pOH$ is given by the Henderson-Hasselbalch equation: $pOH = pK_b + \log \left( \frac{[salt]}{[base]} \right)$.
Since $[B^-] = [HB]$,the ratio $\frac{[salt]}{[base]} = 1$,and $\log(1) = 0$.
Therefore,$pOH = pK_b = -\log(K_b)$.
Given $K_b = 10^{-10}$,we have $pOH = -\log(10^{-10}) = 10$.
We know that $pH + pOH = 14$ at $25^{\circ}C$.
Thus,$pH = 14 - pOH = 14 - 10 = 4$.

For a buffer solution containing equal concentrations of $B^-$ and $HB$,where the $K_b$ of $B$ is $10^{-10}$,what is the $pH$ of the buffer solution?

Similar Questions

Assertion : In a titration of weak acid and $NaOH$,the $pH$ at half equivalence point is $pK_a$.
Reason : At half equivalence point,it forms an acidic buffer and the buffer capacity is maximum where $[acid] = [salt]$.

$A$ solution is prepared by mixing $0.01 \ mol$ each of $H_2CO_3$,$NaHCO_3$,$Na_2CO_3$,and $NaOH$ in $100 \ mL$ of water. The $pH$ of the resulting solution is. . . . . . .
[Given : $pK_{a1}$ and $pK_{a2}$ of $H_2CO_3$ are $6.37$ and $10.32$,respectively; $\log 2=0.30$ ]

An acidic buffer solution can be prepared by mixing a solution of:

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

Which of the following solutions does not act as a buffer?

Vedclass Test Series

Exam Paper Generator

Online Exam Module