Question

Arsenious acid is a weak acid:

HAsO₂<-> H⁺+AsO₂^- pKa=9.22

If 10^-3 moles of the herbicide NaAsO₂ are added to 1 liter of water at 25 degrees Celsius, what is the resulting pH? Determine your answer algebraically using appropriate assumptions or using iterative calculations in a spreadsheet.

Answer 1

As a salt, NaAsO₂ is a strong electrolyte (completely dissociated.

c = [AsO₂] = 1x10^-3 M in this solution.

AsO₂^- is a weak base and its pK_b is

14 – 9.22 = 4.78

Step 1

First use the simplified formula to calculate pOH

[OH-] = (K_b c_base)^1/2

          = (1x10^-4.78 x 1x10^-3) ^1/2

          = (1x10^-7.78 )^1/2

          = 1x10^-7.78

          = 1x10^-3.89

pOH = 3.89

pH    = 14 – pH = 10.11

Step 2 (verification)

Observe that

[OH-] = 1x10^-3.89 = 10^0.11 x 10^-4 = 1.28 x 10^-4 M

In the hydrolysis process

AsO₂^- + H2O   =   HAsO₂ + HO-

[AsO₂^-] = 1x10^-3 - 1.28 x 10^-4 M = 8.72 x10^-4 M

The dissociation rate is 12.8 %, we want it be < 5 % to use the simplified formula.

Than we have to correct c_base to the actual value of the undissociated base [AsO₂^-].

Step 3. Repeat step 1

[OH-] = (K_b c)^1/2

           = (1x10^-4.78 x 8.72 x10^-4 M ) ^½

           = 16.60 x 10^-6 x 8.72 x10^-4 M ) ^½

           = 1.2 x 10^-4M

pOH = 4-0.08 = 3.92

pH = 10.08

Step 4. Verification

dpH = 10.11 – 10.08 = 0.03

Repeat steps 2-4, until dpH<0.02

But you may estimate that the next next result is

pH = 10.07

so you may stop here.

This one is the iteration model.

Alternatively you may use the quadratic formula

[HO-] = x

Kb = x²/(c_base – x)

Solve for x.