Question

100 mL of buffer solution that is 0.1 5M HA (Ka= 6.8x10-5) and 0.20M NaA is mixed with 12.6mL of 0.25M HCI, what is the pH of the resulting solution?

Answer 1

Moles of HA= Molarity*Volume in L= 0.15M*0.1L= 0.015moles

Moles of NaA = 0.20M*0.1L = 0.020moles

Moles of HCl = 0.25M*0.0126L = 0.00315moles

After addition of HCl, HCl reacts completely with the conjugate base, so

HCl + NaA ---> HA + Na⁺

So, moles of NaA = 0.020moles-0.00315moles = 0.0168moles

Moles of HA = 0.015moles + 0.00315moles = 0.0181 moles

Ka = 6.8*10^-5

pKa = -log(6.8*10^-5) = 4.17

Using Henderson-Hasslebalch equation-

pH = pKa + log[NaA]/[HA]

pH = 4.17 + log0.0168/0.0181

= 4.17 - 0.03

pH = 4.137