Question

Benzoic acid has the Ka value= 6.3x10^-5. A 35.0mL sample of 0.13M benzoic acid is titrated w/ a 0.15M solution of NaOH. Find the pH at the equivalence point???

Answer 1

find the volume of kOH used to reach equivalence point

M(C6H5COOH)*V(C6H5COOH) =M(kOH)*V(kOH)

0.13 M *35.0 mL = 0.15M *V(kOH)

V(kOH) = 30.3333 mL

we have:

Molarity of C6H5COOH = 0.13 M

Volume of C6H5COOH = 35 mL

Molarity of kOH = 0.15 M

Volume of kOH = 30.3333 mL

mol of C6H5COOH = Molarity of C6H5COOH * Volume of C6H5COOH

mol of C6H5COOH = 0.13 M * 35 mL = 4.55 mmol

mol of kOH = Molarity of kOH * Volume of kOH

mol of kOH = 0.15 M * 30.3333 mL = 4.55 mmol

We have:

mol of C6H5COOH = 4.55 mmol

mol of kOH = 4.55 mmol

4.55 mmol of both will react to form C6H5COO- and H2O

C6H5COO- here is strong base

C6H5COO- formed = 4.55 mmol

Volume of Solution = 35 + 30.3333 = 65.3333 mL

Kb of C6H5COO- = Kw/Ka = 1*10^-14/6.3*10^-5 = 1.587*10^-10

concentration ofC6H5COO-,c = 4.55 mmol/65.3333 mL = 0.0696M

C6H5COO- dissociates as

C6H5COO- + H2O -----> C6H5COOH + OH-

0.0696 0 0

0.0696-x x x

Kb = [C6H5COOH][OH-]/[C6H5COO-]

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.587*10^-10)*6.964*10^-2) = 3.325*10^-6

since c is much greater than x, our assumption is correct

so, x = 3.325*10^-6 M

[OH-] = x = 3.325*10^-6 M

we have below equation to be used:

pOH = -log [OH-]

= -log (3.325*10^-6)

= 5.48

we have below equation to be used:

PH = 14 - pOH

= 14 - 5.48

= 8.52

Answer: 8.52