# Chapter 15 - Additional Aqueous Equilibria - Problem Solving Practice 15.7 - Page 673: a

$pH = 4.14$

#### Work Step by Step

- Find the numbers of moles: $C(CH_3COOH) * V(CH_3COOH) = 0.1* 0.05 = 5 \times 10^{-3}$ moles $C(NaOH) * V(NaOH) = 0.1* 0.01 = 1 \times 10^{-3}$ moles Write the acid-base reaction: $CH_3COOH(aq) + NaOH(aq) -- \gt NaCH_3COO(aq) + H_2O(l)$ - Total volume: 0.05 + 0.01 = 0.06L Since the base is the limiting reactant, only $0.001$ mol of the compounds will react. Therefore: Concentration (M) = $\frac{n(mol)}{Volume(L)}$ $[CH_3COOH] = 0.005 - 0.001 = 0.004$ moles. Concentration: $\frac{ 0.004}{ 0.06} = 0.06666M$ $[NaOH] = 0.001 - 0.001 = 0$ $[NaCH_3COO] = 0 + 0.001 = 0.001$ moles. Concentration: $\frac{ 0.001}{ 0.06} = 0.01667M$ - Calculate the $pK_a$ for the acid $pKa = -log(Ka)$ $pKa = -log( 1.8 \times 10^{- 5})$ $pKa = 4.745$ - Using the Henderson–Hasselbalch equation: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 4.745 + log(\frac{0.01667}{0.06666})$ $pH = 4.745 + log(0.25)$ $pH = 4.745 + -0.602$ $pH = 4.143$

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