## Chemistry: The Molecular Science (5th Edition)

$4.1g$ of $NaCH_3COO$.
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 $mass(g) = mm(g/mol) * n(moles)$ $mass(g) = 82.04 * 0.05016$ $mass(g) = 4.115$