Question

Consider the titration of 50.0 mL of 0.100 M HOCL (Ka = 3.5 x 10^-8 ) with 0.0400 M NaOH.

a) Calculate how many mL of base are required for the titration to reach equivalent point

b) calculate the pH at the equivalence point.

Answer 1

1)

find the volume of NaOH used to reach equivalence point

M(HOCL)*V(HOCL) =M(NaOH)*V(NaOH)

0.1 M *50.0 mL = 0.04M *V(NaOH)

V(NaOH) = 125 mL

Answer: 125 mL

2)

Given:

M(HOCL) = 0.1 M

V(HOCL) = 50 mL

M(NaOH) = 0.04 M

V(NaOH) = 125 mL

mol(HOCL) = M(HOCL) * V(HOCL)

mol(HOCL) = 0.1 M * 50 mL = 5 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.04 M * 125 mL = 5 mmol

We have:

mol(HOCL) = 5 mmol

mol(NaOH) = 5 mmol

5 mmol of both will react to form OCl- and H2O

OCl- here is strong base

OCl- formed = 5 mmol

Volume of Solution = 50 + 125 = 175 mL

Kb of OCl- = Kw/Ka = 1*10^-14/3.5*10^-8 = 2.857*10^-7

concentration ofOCl-,c = 5 mmol/175 mL = 0.0286M

OCl- dissociates as

OCl- + H2O -----> HOCL + OH-

0.0286 0 0

0.0286-x x x

Kb = [HOCL][OH-]/[OCl-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((2.857*10^-7)*2.857*10^-2) = 9.035*10^-5

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

so, x = 9.035*10^-5 M

[OH-] = x = 9.035*10^-5 M

use:

pOH = -log [OH-]

= -log (9.035*10^-5)

= 4.04

use:

PH = 14 - pOH

= 14 - 4.04

= 9.96

pH = 9.96