In: Chemistry
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?
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