Question

Calculate final pH resulting from the addition of 5.0 mmol of strong acid to the buffer solution made from 0.050 L of 0.25 M NaH2PO4 and 0.100 L of 0.10 M NaOH. pKa(H2PO4-)=7.21.

Answer 1

initially:

mol of H2PO4- ion = MV = 0.25*0.05 = 0.0125 mol of H2PO4-

after addition of mol of NaOH = MV = 0.1*0.1 = 0.01 mol

there i sneutralization

mol of H2PO4- left = 0.0125 -0.010

mol of HPO4-2 formed = 0.010 mol

then, there is addition of

5 mmol = 5*10^-3 = 0.005 mol of acid

mol of H2PO4- left = 0.0125 -0.010 + 0.005 = 0.0075

mol of HPO4-2 formed = 0.010 -0.005 = 0.005

this is a buffer so

pH = pKa2 ´log/HPO4-2 / H2PO4-)

pH = 7.21 + log(0.005/0.0075)=7.0339

pH = 7.0339