Question

Calculate the mass of ${NH}_{4}Cl$ dissolved in $500ml$ to have $pH=4.5$. ($K_b$ of ${NH}_{4}OH=1.8\times {10}^{-5})$

• A. $47.76g$
• B. $2\times {10}^{-5}g$
• C. $5.35g$
• D. $53.5g$

Answer 1

Answer: (A)

$47.76g$

As $N{H}_{4}Cl$ is a salt of weak base & strong acid,

$pH=\frac{1}{2}pKw-\frac{1}{2}\log c-\frac{1}{2}pKb\therefore 4.5=\frac{1}{2}\times 14-\frac{1}{2}\log c-\frac{1}{2}\times \log (1.8\times {10}^{-5})\therefore \log c=0.26$

But $C=\frac{x}{26.25}$

where $x$= amount of $N{H}_{4}Cl$ dissolved in $500ml$ of water

$\therefore \log c=\log \left(\frac{x}{26.25}\right)\therefore 0.26=\log \left(\frac{x}{26.25}\right)\therefore x=47.76g$