Question

Calculate the pH after 10.0 mL of 0.400 M NaOH is added to 20.0 mL of 0.50 M CH₃COOH.(Ka of CH₃COOH is 1.8 * 10⁻⁵)

The answer is supposed to be 4.57, but I'm getting 4.37. I'm so close but need a little guidance.

Answer 1

pH = pKa + log [Salt]/[acid]

moles of NaOH = 0.01L x 0.4 M = 0.004 moles

moles of CH3COOH = 0.02 L x 0.5M = 0.01 moles

NaOH + CH₃COOH ---------> CH₃COONa + H₂O

0.004 0.01 0 - initially

0 0.01 - 0.004 0.004 - after reaction

total volume = 0.01 L + 0.02L = 0.03 L

salt = CH₃COONa = 0.004 moles, molarity = moles /volume =  0.004 / 0.03

acid remaining = 0.01 - 0.004 = 0.006 moles , molarity = 0.006 /0.03

pKa = -log(Ka) = -log(1.8 x 10^-5) = 4.7447

log[salt]/[acid] = log[0.004/.03]/[0.006/.03] = -0.176

pH = 4.7447 - 0.176 = 4.5686 ~ = 4.57