Question

Consider the titration of 40.0 mL of 0.250 M HF with 0.200 M NaOH. How many milliliters of base are required to reach the equivalence point? Calculate the pH at the following points:

a)After the addition of 10.0 mL of base

b)Halfway to the equivalence point

c)At the equivalence point

d)After the addition of 80.0 mL of base

Answer 1

no of mol of HF = 40/1000*0.25 = 0.01 mol

pka of HF = 3.17

a)After the addition of 10.0 mL of base

NO of mol of NaOH added = 10/1000*0.2 = 0.002 mol

pH = pka + log(salt/acid)

    = 3.17+log(0.002/(0.01-0.002))

     = 2.57

b)Halfway to the equivalence point , the pH of buffer = pka of acid

= 3.17

c)At the equivalence point

No of mol of HF = NO of mol of NaOH added

concentration of salt = 0.01/(0.05+0.04) = 0.11 M

pH = 7+1/2(pka + logC)

    7+1/2(3.17+log0.11) = 8.1

d)After the addition of 80.0 mL of base

No of mol of base added = 80/1000*0.2 = 0.016 mol

excess No of mol of base added 0.016 - 0.01 = 0.006 mol

concentration of excess base = 0.006 / 0.12 = 0.05 M

pOH = -log(OH-)

     = -log0.05 = 1.3

pH = 14-1.3 = 12.7