# Chapter 15 - Additional Aqueous Equilibria - Questions for Review and Thought - Topical Questions - Page 693a: 21

We should add 1.5 grams of $NH_4Cl$ to form that buffer.

#### Work Step by Step

1. Since $N{H_4}^+$ is the conjugate acid of $NH_3$ , we can calculate its ka by using this equation: $K_b * K_a = K_w = 10^{-14}$ $1.8\times 10^{- 5} * K_a = 10^{-14}$ $K_a = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$ $K_a = 5.556\times 10^{- 10}$ 2. Calculate $[H_3O^+]$: $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 9.00}$ $[H_3O^+] = 1.000 \times 10^{- 9}$ 3. Write the $K_a$ equation, and find the acid concentration: $K_a = \frac{[H_3O^+][N{H_4}^+]}{[NH_3]}$ $5.556 \times 10^{-10} = \frac{1 \times 10^{-9}*[N{H_4}^+]}{[0.1]}$ $5.556 \times 10^{-10} = 1 \times 10^{-8}*[N{H_4}^+]$ $0.05556M = [N{H_4}^+]$ 4. Calculate the number of moles: $n(moles) = concentration(M) * volume(L)$ $n(moles) = 0.05556 * 0.5$ $n(moles) = 0.02778$ 5. Find the mass value in grams: 14.01* 1 + 1.01* 4 + 35.45* 1 = 53.5g/mol $mass(g) = mm(g/mol) * n(moles)$ $mass(g) = 53.5 * 0.02778$ $mass(g) = 1.486$

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