Question

Using the aqueous equilibrium constants, calculate the concentration of OH- in M and pH for the following solutions (given that all are at 25 degrees celsius):

0.15 M NaBrO

8.2×10⁻² M NaHS.

A mixture that is 0.13 M in NaNO2 and 0.25 M in Ca(NO2)2.

Thank you! :)

Answer 1

1) 0.15 M NaBrO

HBrO , Ka value = 2.8 x 10^-9

Kb = Kw / Ka = 1.0 x10^-14 / 2.8 x 10^-9

Kb = 3.57 x 10^-6

BrO-     + H2O <-------------------> HBrO   + OH-

0.15                                                 0             0 -------------> initial

0.15-x                                             x              x --------------> equilibrium

Kb = [HBrO][OH-]/[BrO-]

3.57 x 10^-6 = x^2 / 0.15-x

x^2 + 3.57 x 10^-6 x - 5.36 x 10^-7 = 0

x = 7.30 x 10^-4

[OH-] = x = 7.30 x 10^-4 M

pOH = -log [OH-] = -log (7.30 x 10^-4 ) = 3.14

pH + pOH = 14

pH = 10.86

2) 8.2×10⁻² M NaHS.

H2S , Ka value = 8.9 x 10^-9

Kb = 1.12 x 10^-6

HS-   + H2O ----------------------> H2S + OH-

8.2 x 10^-2 -x                           x              x ---------------> equilibrium

Kb = x^2 / 8.2 x 10^-2 -x  

1.12 x 10^-6 = x^2 / 8.2 x 10^-2 -x  

x^2 + 1.12 x 10^-6 x - 9.21 x 10^-8 = 0

x = 3.03 x 10^-4

[OH-] = x = 3.03 x 10^-4 M

pOH = -log[OH-] = -log (3.03 x 10^-4 ) = 3.52

pH + pOH = 14

pH = 10.48