Question

You are titrating 0.200 L of a 0.400M monoprotic weak acid with a strong base that is 0.800 M.

Ka= 4.8x10^-6

a) At first, there is 0.200 L of 0.400M acid and no strong base. What is the pH?

b) After 35.0 mL of 0.800 M strong base is added, what is the pH?

c) How many mL of 0.800 M strong base must be added to reach the half equivalence point?

d) What is the pH of the equivalence point?

f) What is the pH after 102mL of 0.800 M strong base is added?

g) If the strong base were NaOH, how many moles of NaOH is present at the equivialence point, half-equivalences point and buffer region?

Answer 1

pka of acid = -log(4.8*10^(-6)) =5.32

No of mol of acid = 0.2*0.4 = 0.08 mol

a) pH = 1/2(pka-logC)

      = 1/2(5.32-log0.08)

      = 3.2

b) No of mol of base = 35/1000*0.8 = 0.028 mol

pH = pka + log(salt/base)

     = 5.32 + log(0.028/(0.08-0.028))

     = 5.05

c) at half equivalent No of mol of base = 1/2*0.08 = 0.04 mol

   volume of base = 0.04/0.8 = 0.05 L

    = 50 ml

e) at equivalencepoint

concentration of salt = 0.08/(0.2+0.1) = 0.27 M

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

    = 7+1/2(5.32+0.27)
    = 9.8

f) No of mol of base = 0.102*0.8 = 0.0816 mol

   no of mol of acid = 0.08 mol

excess base = 0.0816-0.08 = 0.0016 mol

concentration of base = 0.0016/(0.2+0.102) = 0.0053 M

POH = -LOG0.0053 = 2.27

Ph = 14-2.27 = 11.73