Answer
The necessary mass of ammonium chloride is equal to 0.48g.
Work Step by Step
1. Since ${NH_4}^+$ is the conjugate acid of $NH_3$ , we can calculate its $K_a$ by using this equation:
$K_b * K_a = K_w = 10^{-14}$
$ 1.8\times 10^{- 5} * K_a = 10^{-14}$
$K_a = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$
$K_a = 5.6\times 10^{- 10}$
2. Calculate $[H_3O^+]$:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 9}$
$[H_3O^+] = 1 \times 10^{- 9}$
3. Write the $K_a$ equation, and find the needed acid concentration:
$K_a = \frac{[H_3O^+][NH_3]}{[N{H_4}^+]}$
$5.6 \times 10^{-10} = \frac{1 \times 10^{-9}*[0.1]}{[N{H_4}^+]}$
$N{H_4}^+ = \frac{(0.1) * (1 \times 10^{-9}) }{[5.6 \times 10^{-10}]}$
${[N{H_4}^+]} = 0.18$
4. Find the number of moles:
$Concentration(M) = \frac{n(mol)}{V(L)}$
$0.18 = \frac{n(mol)}{0.05}$
$0.18 * 0.05 = n(mol)$
$8.9 \times 10^{-3} moles = n(mol)$
5. Determine the molar mass of this compound ($NH_4Cl$):
14.01* 1 + 1.01* 4 + 35.45* 1 = 53.5g/mol
6. Calculate the mass
$mm(g/mol) = \frac{mass(g)}{n(mol)}$
$mm(g/mol) * n(mol) = mass(g)$
$ 53.5 * 8.9 \times 10^{-3} = mass(g)$
$0.48 = mass(g)$
