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

$pH = 5.39$
1. Calculate the molar mass $(NaCH3COOH)$: 22.99* 1 + 12.01* 1 + 1.01* 3 + 12.01* 1 + 16* 1 + 16* 1 + 1.01* 1 = 83.05g/mol 2. Calculate the number of moles $(NaCH3COOH)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 83}{ 83.05}$ $n(moles) = 1$ 3. Find the concentration in mol/L $(NaCH3COOH)$: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 1}{ 1.5}$ $C(mol/L) = 0.67$ 4. Calculate the pKa Value $pKa = -log(Ka)$ $pKa = -log( 1.8 \times 10^{- 5})$ $pKa = 4.74$ 5. Check if the ratio is between 0.1 and 10: - $\frac{[Base]}{[Acid]} = \frac{0.67}{0.15}$ - 4.4: It is. 6. Check if the compounds exceed the $K_a$ by 100 times or more: - $\frac{0.67}{1.8 \times 10^{-5}} = 3.7\times 10^{4}$ - $\frac{0.15}{1.8 \times 10^{-5}} = 8333$ 7. Using the Henderson–Hasselbalch equation: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 4.74 + log(\frac{0.67}{0.15})$ $pH = 4.74 + log(4.4)$ $pH = 4.74 + 0.65$ $pH = 5.39$