Question

Two 25.0 ml samples, one 0.100 mol/L HCl and the other 0.100mol/L HF, were titrated with 0.200 mol/L KOH.

a)What is the volume of added base at the equivalence point for each titration?

b)Predict whether the pH at the equivalence point for each titration will be acidic, basic or neutral? Explain your selection.c)Predict which titration curve will have the lower initial pH and explain your selection.

Answer 1

2 samples

V = 25 ml each

M = 0.1 M of HCl

M = 0.1 M of HF

with

M = 0.2 M o fKOH

a)

the equivalence point must be the same since

mol of acid = mol of base

mol of acid = 25*0.1 = 2.5 mmol of acid

then we need 2.5 mmol of base

mmol of base = M*V

2.5 = 0.2*V

V = 2.5/0.2 = 12.5 ml of base

same case for HF and HCl

b)

pH in equivalencepoint

for HCl ---> strong acid + strong bases = pH must be ALWAYS neutral, since it forms a neutral base + water

HCl + KOH = H2O + KCl

for

HF

it forms

HF + KOH = H2O + KF

F- ions will be insolution so

F- + H2O = HF + OH-

then

Kb = [HF][OH-]/[F-]

Kb = Kw/Ka

and Ka = 7.2*10^-4

so

Kb = (10^-14)/(7.2*10^-4) = 1.388*10^-11

therefore

Kb = [HF][OH-]/[F-]

HF = OH = x since we dont know

F- = mol of F-/ total volume = 2.5/(25+12.5)

[F-] = 0.06666666 - x (x is the dissoaciated acid in equilibrium)

then

1.388*10^-11 = x*x/(0.0666)

solve for x

x = sqrt((1.388*10^-11)(0.066)) = 9.5712068*10^-7

x = Oh- = 9.5712068*10^-7

pOH = -log(9.5712068*10^-7) = 6.02

pH = 14-6.02 = 7.98

c)

the titration curve will be:

for HCL

no changes in pH and then suddenly, in the equivalence point; a drastic change of pH (i.e. from acidic to basic)

for HF

slightly changes of pH; then buffer formation, pH = pKa in the half equivalence point; then a large change of pH in equivalenc point (not equal to 7)