Answer
14.8 g of $NaCH_3COO$ in order to get a 0.100 M $CH_3COOH$ solution to pH = 5.00
Work Step by Step
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ CH_3COOH ]& [ CH_3COO^- ]& [ H_3O^{+} ]\\
Initial& 0.100 & y & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.100 -x& y +x& 0 +x\\
\end{vmatrix}$$
Where "y" is the amount of $NaCH_3COO$ we must add to get pH = 5.00
$$[H_3O^+] = 10^{-pH} = 10^{-5.00} = 1.0 \times 10^{-5} \space M$$
$x = [H_3O^+] = 1.0 \times 10^{-5} \space M$
Substituting into the Ka expression:
$$K_a = 1.8 \times 10^{-5} = \frac{(y + 1.0 \times 10^{-5})(1.0 \times 10^{-5})}{0.100 - 1.0 \times 10^{-5}}$$
Solving for y:
y = 0.180
2. Now, find how many grams we need to get this molarity:
$ NaCH_3COO $ : ( 22.99 $\times$ 1 )+ ( 1.008 $\times$ 3 )+ ( 12.01 $\times$ 2 )+ ( 16.00 $\times$ 2 )= 82.03 g/mol
- Use all the information as conversion factors:
$$ 1.00 \space L \times \frac{ 0.180 \space mol}{1 \space L} \times \frac{ 82.03 \space g}{1 \space mol} = 14.8 \space g$$
