Question

Calculate the pH of the following:

A. 10. 0 ml of 0.100 M HC₆H₅O₂ (K_a = 6.5 x 10^-5) and 10.0 ml of 0.10 M NaOH

B. 25.0 mL of 0.1 M HC₆H₅O₂ (K_a = 6.5 x 10^-5) and 15.0 mL of 0.1 M NaOH

Answer 1

For the case A.

pH = pKa + log[S]/[A] pKa = -log(6.5x10^-5) = 4.187

moles of acyd = 0.1x0.010 = 0.001 moles moles of NaOH = 0.1x0.010 = 0.001 moles

at the end of the reaction, we have 20 mL of solution, so the molar concentration of the salt would be: 0.001/0.020 = 0.05 M

pH = 4.187 + log [0.05] / [0.1] = 4.187 - 0.3 = 3.885

For the case B.

pH = pKa + log[S]/[A] pKa = -log(6.5x10^-5) = 4.187

moles of acyd = 0.1x0.025 = 0.0025 moles moles of NaOH = 0.1x0.015 = 0.0015 moles

at the end of the reaction, we have 40 mL of solution, so the molar concentration of the salt would be: 0.0015/0.040 = 0.0375 M

pH = 4.187 + log [0.0375] / [0.1] = 4.187 - 0.3 = 3.761