Question

500.0 mL of 0.100 M NaOH is added to 535 mL of 0.250 M weak acid (Ka = 7.69 × 10-5). What is the pH of the resulting buffer?

Answer 1

Volume of 0.100 M NaOH = 500.0 mL = 0.500 L

Volume of weak acid = 535 mL = 0.535 L , Molarity of weak acid = 0.250

Ka of weak acid = 7.69 E -5

Lets calculate moles of NaOH

Mol NaOH = 0.500 L * 0.100 M

= 0.050 mol

Mol weak acid = 0.250 M * 0.535 L

= 0.13375 mol

NaOH reacts with weak acid

Lets assume weak acid is HA

Reaction between HA and NaOH

NaOH (aq)   + HA (aq)   --- > NaA (aq)   + H2O (l)

I           0.050              0.13375           0                      0

C          -0.050              -0.050              +0.050            

E          0                      0.08375                       0.050  

Molarity of

[HA] = 0.08375 mol / total volume = 0.08375 mol / ( 0.500 + 0.535 ) L

= 0.081 M

[NaA] = [ A-]= 0.050 mol/ (0.500 + 0.535)

=0.048309 M

We now use Henderson Hassellbalch equation

pH = pka + log ([A- ] / [ HA])

            pka = -log ka

                    = - log ( 7.69 E-5)

                    = 4.11

pH = 4.11 + log ( 0.048309 / 0.081 )

= 3.89

pH of the buffer = 3.89