Question

The number of grams/weight of $NH_4Cl$ required to be added to 3 liters of $0.01\, M\, NH_3$ to prepare the buffer of $pH=9.45$ at temperature $298\, K$ ($K_b$ for $NH_3$ is $1.85 \times 10^{-5}$)

• A. 3.53 gm
• B. 0.354 gm
• C. 4.55 gm
• D. 0.455gm

Answer 1

Answer: (B)

0.354 gm

Answer 2

$\because pOH = pK _{ b }+\log \frac{\left[ NH _{4} Cl \right]}{\left[ NH _{3}\right]}$
$pOH =-\log K_{b}+\log \frac{\left[ NH _{4} Cl \right]}{\left[ NH _{3}\right]}$
$\left(\because p K_{b}=-\log K_{b}\right)$

Also, $pOH =14- pH =14-9.45=4.55$

and $\log \frac{\left[ NH _{4} Cl \right]}{\left[ NH _{3}\right] \cdot K_{b}}=\log 3 \approx 0.470$

Thus, $\log 10^{14}-\left(\log 10^{9}+\log 3\right) \approx \log \frac{\left[ NH _{4} Cl \right]}{\left[ NH _{3}\right] \cdot K_{b}}$
$\Rightarrow \, \log \frac{10^{14}}{10^{9} \times 3}=\log \frac{\left[ NH _{4} Cl \right]}{\left[ NH _{3}\right] \cdot K_{b}}$
$\Rightarrow \frac{10^{14}}{10^{9} \times 3} \approx \frac{\left[ NH _{4} C l\right]}{\left[ NH _{3}\right] \cdot K_{b}}$

or $\frac{10^{14} \times\left[ NH _{3}\right] \cdot K_{h}}{10^{9} \times 3}$
$=\left[ NH _{4} Cl \right] \approx \frac{10^{14}}{10^{9} \times 3} \times 0.01 \times 1.85 \times 10^{-5}$
$=\left[ NH _{4} Cl \right] \approx 0.354\,gm$