Question

To prepare a buffer solution of $pH=4.04$, amount of Barium acetate to be added to $100\text{mL}$ of $0.1M$ acetic acid solution $\left[p{K}_b\right(C{H}_{3}CO{O}^{-})=9.26]$ is:
Given ${10}^{-0.7}\approx 0.2$

• A. 0.05 mole
• B. 0.025 mole
• C. 0.1 mole
• D. 0.005 mole

Answer 1

The correct option is B 0.025 mole

Buffer solution is the mixture of weak acid and its conjugate base.

For the buffer solution of $C{H}_{3}COOH$ and $C{H}_{3}CO{O}^{-}$,
$pH=p{K}_a+\text{log}\frac{\left[C{H}_{3}COOH\right]}{\left[C{H}_{3}CO{O}^{-}\right]}(byHendersonHasselbalchequation$
Given,
$p{K}_b\left(C{H}_{3}CO{O}^{-}\right)=9.26$
$p{K}_a\left(C{H}_{3}CO{O}^{-}\right)=14-9.26=4.74$
Concentration of acetic acid = 0.1 M
Millimoles of acetic acid = 0.1 $\times$ 100 = 10 mmoles
$\Rightarrow 4.04=4.74+\text{log}\frac{(100\times 0.1)}{\text{m. moles of}C{H}_{3}CO{O}^{-}}$


$\Rightarrow$ m. moles of $C{H}_{3}CO{O}^{-}=50$
Moles of $C{H}_{3}CO{O}^{-}=0.05$
$\therefore$ moles of $(C{H}_{3}COO{)}_{2}Ba$ required $=0.025$