Answer
The mass of sodium acetate needed is equal to 4.7g.
Work Step by Step
1. Calculate $[H_3O^+]$:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 4.5}$
$[H_3O^+] = 3.16 \times 10^{- 5}M$
2. Write the $K_a$ equation, and find the ratio:
$K_a = \frac{[H_3O^+][CH_3C{O_2}^-]}{[CH_3C{O_2}H]}$
$1.8 \times 10^{-5} = \frac{3.16 \times 10^{-5}*[CH_3C{O_2}^-]}{[CH_3C{O_2}H]}$
$\frac{1.8 \times 10^{-5}}{3.16 \times 10^{-5}} = \frac{[CH_3C{O_2}^-]}{[CH_3C{O_2}H]}$
$0.57 = \frac{[CH_3C{O_2}^-]}{[CH_3C{O_2}H]}$
3. Calculate the needed concentration of the base:
$0.57 = \frac{[CH_3C{O_2}^-]}{0.10}$
$0.57 * 0.10 = {[CH_3C{O_2}^-]}$
$0.057M = {[CH_3C{O_2}^-]}$
** Notice: $[NaCH_3C{O_2}^-] = [CH_3C{O_2}^-] = 0.057M$
4. Find the number of moles:
$Concentration(M) = \frac{n(mol)}{V(L)}$
$0.057 = \frac{n(mol)}{1}$
$0.057 * 1 = n(mol)$
$0.057 moles = n(mol)$
5. Determine the molar mass of this compound $(NaCH_3CO_2)$:
** Remember: we are adding sodium acetate, not just the base ion.
22.99* 1 + 12.01* 1 + 1.01* 3 + 12.01* 1 + 16* 2 = 82.04g/mol
6. Calculate the mass
$mm(g/mol) = \frac{mass(g)}{n(mol)}$
$mm(g/mol) * n(mol) = mass(g)$
$ 82.04 * 0.057 = mass(g)$
$4.7 = mass(g)$
