Question

What is the pH if 2ml of 2M HCl is added to 100ml of a solution that is .25M CH3COOH and .35M NaCH3COO?

Answer 1

You forgot to put the value of Ka or pKa. However the reported values in general for this is 1.75x10^-5 so pKa = -log(1.75x10^-5) = 4.76

Now the overall reaction:

CH₃COOH = CH₃COO^- + H⁺

Adding acid means that the concentration of H⁺ is increasing, and concentration of acetate ion is decreasing so:

pH = pKa + log(CH₃COO^- - H⁺ / CH₃COOH + H⁺)

moles of acid = 2 * 0.002 = 0.004 moles

moles of acetic acid = 0.25 * 0.100 = 0.025 moles

moles of acetate = 0.35 * 0.100 = 0.035 moles

we can neglect the total volume of 102 mL so:

pH = 4.76 + log(0.035-0.004 / 0.025+0.004)

pH = 4.76 + 0.028963

pH = 4.79

Hope this helps