91影视

Calculate the number of moles of benzoic acid. The molar mass of benzoic acid (C鈧嘓鈧哋鈧�) is approximately 122.12 g/mol. Use the formula: \( \, \text{moles} = \frac{\text{mass}}{\text{molar mass}} \, \).\[ \text{moles of benzoic acid} = \frac{1.24 \, \text{g}}{122.12 \, \text{g/mol}} \approx 0.0102 \, \text{mol} \]

Using the stoichiometry of the titration reaction, determine how much NaOH is needed to neutralize the benzoic acid. The reaction between NaOH and benzoic acid is 1:1, so the moles of NaOH required are 0.0102 mol.Using the concentration of NaOH:\[ V = \frac{\text{moles}}{\text{concentration}} = \frac{0.0102 \text{ mol}}{0.180 \text{ M}} \approx 0.0567 \text{ L} \approx 56.7 \text{ mL} \]

Since 56.7 mL of 0.180 M NaOH is exactly sufficient to reach the equivalence point of the titration, we know the reaction is complete. At the equivalence point, all of the benzoic acid has reacted with NaOH, forming its conjugate base, which affects the pH.

At equivalence, the only species affecting the pH is the conjugate base of the benzoic acid, the benzoate ion (C鈧咹鈧匔OO鈦�).The concentration of benzoate ions can be calculated by considering the total volume: initial 50 mL + NaOH 56.7 mL = 106.7 mL.\[ \text{Concentration of C鈧咹鈧匔OO}^- = \frac{0.0102 \text{ mol}}{0.1067 \text{ L}} \approx 0.0957 \text{ M} \]Calculate the pH using the formula involving K_b of benzoate and its concentration.Given: \( K_a \text{ of benzoic acid} \approx 6.3 \times 10^{-5} \), so \( K_b = \frac{K_w}{K_a} \approx \frac{1.0 \times 10^{-14}}{6.3 \times 10^{-5}} \approx 1.59 \times 10^{-10} \).Using the formula \( [OH鈦籡 = \sqrt{K_b \times [\text{C鈧咹鈧匔OO}^-] } \), we find:\[ [OH鈦籡 = \sqrt{1.59 \times 10^{-10} \times 0.0957} \approx 1.23 \times 10^{-5} \, \text{M} \]Convert [OH鈦籡 to pOH and then to pH:\[ \text{pOH} = -\log(1.23 \times 10^{-5}) \approx 4.91 \]\[ \text{pH} = 14 - \text{pOH} = 14 - 4.91 = 9.09 \]