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

$pH = 8.31$
- Find the numbers of moles: $C(C_6H_5COOH) * V(C_6H_5COOH) = 0.1* 0.03 = 3 \times 10^{-3}$ moles $C(NaOH) * V(NaOH) = 0.1* 0.03 = 3 \times 10^{-3}$ moles When the number of moles is equal, the reactants are totally consumed: $C_6H_5COOH(aq) + NaOH(aq) -- \gt NaC_6H_5COO(aq) + H_2O(l)$ - Total volume: 0.03 + 0.03 = 0.06L - So, these are the final concentrations: $[C_6H_5COOH] = 0.003 - 0.003 = 0$ mol. $[NaOH] = 0.003 - 0.003 = 0$ mol $[NaC_6H_5COO] = 0 + 0.003 = 0.003$ moles. Concentration: $\frac{ 0.003}{ 0.06} = 0.05M$ - Therefore, we have a weak base salt solution: - Since $C_6H_5COO^-$ is the conjugate base of $C_6H_5COOH$, we can calculate its kb by using this equation: $K_a * K_b = K_w = 10^{-14}$ $1.2\times 10^{- 4} * K_b = 10^{-14}$ $K_b = \frac{10^{-14}}{ 1.2\times 10^{- 4}}$ $K_b = 8.333\times 10^{- 11}$ - We have these concentrations at equilibrium: -$[OH^-] = [C_6H_5COOH] = x$ -$[C_6H_5COO^-] = [C_6H_5COO^-]_{initial} - x = 0.05 - x$ For approximation, we consider: $[C_6H_5COO^-] = 0.05M$ - Now, use the Kb value and equation to find the 'x' value. $Kb = \frac{[OH^-][C_6H_5COOH]}{ [C_6H_5COO^-]}$ $Kb = 8.333 \times 10^{- 11}= \frac{x * x}{ 0.05}$ $Kb = 8.333 \times 10^{- 11}= \frac{x^2}{ 0.05}$ $4.167 \times 10^{- 12} = x^2$ $x = 2.041 \times 10^{- 6}$ Percent ionization: $\frac{ 2.041 \times 10^{- 6}}{ 0.05} \times 100\% = 0.004082\%$ %ionization < 5% : Right approximation. Therefore: $[OH^-] = [C_6H_5COOH] = x = 2.041 \times 10^{- 6}M$ $[C_6H_5COO^-] \approx 0.05M$ $pOH = -log[OH^-]$ $pOH = -log( 2.041 \times 10^{- 6})$ $pOH = 5.69$ $pH + pOH = 14$ $pH + 5.69 = 14$ $pH = 8.31$