Answer
$4.1g$ of $NaCH_3COO$.
Work Step by Step
1. Calculate $[H_3O^+]$:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 4.57}$
$[H_3O^+] = 2.692 \times 10^{- 5}$
2. Write the $K_a$ equation, and find the base concentration:
$K_a = \frac{[H_3O^+][NaCH_3COO]}{[CH_3COOH]}$
$1.8 \times 10^{-5} = \frac{2.692 \times 10^{-5}*[NaCH_3COO]}{[0.15]}$
$1.8 \times 10^{-5} = 1.794 \times 10^{-4}*[NaCH_3COO]$
$0.1003M = [NaCH_3COO]$
3. Calculate the number of moles:
$n(moles) = concentration(M) * volume(L)$
$n(moles) = 0.1003 * 0.5$
$n(moles) = 0.05016$
4. Find the mass value in grams:
22.99* 1 + 12.01* 1 + 1.01* 3 + 12.01* 1 + 16* 1 + 16* 1 = 82.04g/mol
$mass(g) = mm(g/mol) * n(moles)$
$mass(g) = 82.04 * 0.05016$
$mass(g) = 4.115$
