A 35.00 mL solution of 0.25M HF is titrated with a standardized 0.1281M solution of NaOH...

A 35.00 mL solution of 0.25M HF is titrated with a standardized 0.1281M solution of NaOH at 25C.
a.What is the PH of the HF solution before titrant is added?
b.How many milliliters of titrant are required to reach the equivalence point?
c.What is the PH at 0.5ml before the equivalence point?
d.What is the PH at the equivalence point?
e.What is the PH at 0.5 ml after the equivalence point?

Solutions

Expert Solution

a. pka of weakacid = 3.17

pH of weakacid = 1/2(pka-logC)

C = concentration of weakacid = 0.25 M

= 1/2(3.173-log0.25)

= 1.89

b. NaOH + HF ----> NaF + H2O

no of mole of HF = 35*0.25/1000 = 0.00875

no of mole of NaOH = 0.00875

volume of NaOH = n/M = 0.00875/0.128 = 0.06836 L

= 68.36 ml
c. before equivalence point

no of mole of HF = 35*0.25/1000 = 0.00875

no of mole of NaOH = 67.86*0.128/1000 = 0.00869

pH = pka + log(salt/acid)

   = 3.17 + log(0.00869/(0.00875-0.00869))

   = 5.33

d. at equivalence point

no of mole of HF = 35*0.25/1000 = 0.00875

no of mole of NaOH = 0.00875

concentration of salt = (0.00875/(68.36+35))*1000 = 0.0846 M

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

= 7+1/2(3.17+log0.0846)

= 8.04

e. after equivalence point

excess NaOH concentration = (0.5*0.1281/(35+68.86))*1000 = 0.6167 M

pH = 14 - (-log(OH-))

= 14 - (-log0.6167)

= 13.8

queen_honey_blossom answered 1 year ago

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