Question

1. Calculate the pH of a solution that results from mixing 15 mL of 0.13 M HBrO(aq) with 11 mL of 0.1 M NaBrO(aq). The K_a value for HBrO is 2 x 10^-9.

2. What is the buffer component ratio, (BrO^-)/(HBrO) of a bromate buffer that has a pH of 7.91. K_a of HBrO is 2.3 x 10^-9.

Answer 1

1)

Lets find the concentration after mixing for 1st component

Concentration after mixing = mol of component / (total volume)

M(HBrO) after mixing = M(HBrO*)V(HBrO)/(total volume)

M(HBrO) after mixing = 0.13 M*15.0 mL/(15.0+11.0)mL

M(HBrO) after mixing = 0.075 M

Lets find the concentration after mixing for 2nd component

Concentration after mixing = mol of component / (total volume)

M(NaBrO) after mixing = M(NaBrO*)V(NaBrO)/(total volume)

M(NaBrO) after mixing = 0.1 M*11.0 mL/(11.0+15.0)mL

M(NaBrO) after mixing = 0.0423 M

Ka = 2*10^-9

pKa = - log (Ka)

= - log(2*10^-9)

= 8.699

use:

pH = pKa + log {[conjugate base]/[acid]}

= 8.699+ log {0.0423/0.075}

= 8.45

Answer: 8.45

2)

Ka = 2*10^-9

pKa = - log (Ka)

= - log(2*10^-9)

= 8.699

use:

pH = pKa + log {[conjugate base]/[acid]}

= 8.699+ log {0.0423/0.075}

= 8.4502

use:

pKa = -log Ka

= -log (2.3*10^-9)

= 8.6383

use:

pH = pKa + log {[conjugate base]/[acid]}

7.91 = 8.6383+log {[BrO-]/[HBrO]}

log {[BrO-]/[HBrO]} = -0.7283

[BrO-]/[HBrO] = 0.187

Answer: 0.187