## Chemistry and Chemical Reactivity (9th Edition)

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 $mm(g/mol) = \frac{mass(g)}{n(mol)}$ $mm(g/mol) * n(mol) = mass(g)$ $53.5 * 8.9 \times 10^{-3} = mass(g)$ $0.48 = mass(g)$