Question

Use what you learned about [H3O(+)] and pH formulas to solve the following problem

Calculate the pH of a (1.99x10^-1) M solution of Na2S03 (Kb1 = 1.56E-7, Kb2 = 5.88E-13)

Answer 1

Solution :-

Na2SO3 is the salt therefore it will form ions as 2 Na+ and SO3^2-

The SO3^2- ion will acts as base when added to water as follows

     SO3^2- + H2O   ----- > HSO3- + OH-

I   0.199                                   0              0

C   -x                                      +x             +x

E 0.199-x                                x               x

So now lets calculate the concentration of the OH- using the kb1

Kb1 = [HSO3-] [OH-] /[SO3^2-]

1.56E-7 = [x][x]/0.199-x]

Since kb value is very small therefore we can neglect the x from the denominator then we get

1.56E-7 = [x][x]/0.199]

1.56E-7 *0.199 =x²

3.104E-8 = x²

1.76E-4 = x

Now lets calculate the concentration of the OH- using the second kb value.

     HSO3- + H2O   ----- > H2SO3 + OH-

I   1.76E-4                               0              0

C   -x                                      +x             +x

E 1.76E-4-x                            x               x

Kb2= [H2SO3] [OH-] /[HSO3-]

5.88E-13= [x][x]/1.76E-4-x]

Neglect the x from denominator because kb2 is very small

Then we get

5.88E-13= [x][x]/1.76E-4]

5.88E-13*1.76E-4= x²

1.03E-16 =x2

Taking square root of both side we get

1.01E-8 = x =[OH-]

Therefore the total [OH-] concentration = 1.76E-4 +1.01E-8 = 1.76 E-4 M

Now lets calculate pOH using this concentration

pOH= -log [OH-]

pOH = - log [1.76 E-4 ]

pOH = 3.75

now using the pOH lets calculate the pH

pH + pOH = 14

pH = 14 – pOH

pH = 14-3.75

pH =10.25