A 40.00 ml sample of 0.10 m weak acid with ka of 1.8*10-5 is titrated with a 0.10 m strong base. what is the ph after 20.00 ml of base has been added?

The answer is 4.74.

Solution:

The product of the volume and concentration of each solution gives the number of moles of the acid HA and the base OH - present before the neutralization:

0.0400L (0.10mol/L HA) = 0.00400 moles HA

0.0200L (0.10mol/L OH-) = 0.00200 moles OH-

We can see that the 0.00200 moles of the strong base OH - consumes 0.00200 moles of the weak acid HA:

HA + OH - → A - + H2O

initial 0.00400 0.00200 0

change - 0.00200 - 0.00200 + 0.00200

equilibrium 0.00200 0 0.00200

The concentration of HA is equal to the concentration of A - after the reaction:

[HA] = [A-] = 0.00200mol / 0.0400L+0.0200L = 0.03333 M

The equilibrium-constant expression for HA is

Ka = [H+][A-] / [HA]

Ka = [H+] = 1.8x10^-5

We can now solve for pH:

pH = - log [H+] = - log (1.8x10^-5) = 4.74