Answer
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
$mass(g) = mm(g/mol) * n(moles)$
$mass(g) = 53.5 * 0.02778$
$mass(g) = 1.486$
