Question

a) Calculate the pH in 0.030M H2SO3(Ka1=1.5

Answer 1

a)

M = 0.030 M of H2SO3

H2SO3 <-> H+ HSO3

HSO3- <-> H+ SO3^-2

This will be formed...

We need the [H+] concentration

From the first equilbrium

Ka1 = [HSO3-][H+] /[H2SO3]

Ka2 = [SO3^-2 ][H+]/ [HSO3-]

There will be two H+ to account, the Ka1 and form Ka2

Let us calculate first the ammount of Ka1 H+ ions

NOTE: [HSO3-] = [H+] = x since for 1 mol of [H+] you will have always another mol of HSO3-

[H2SO3] = 0.030 - x

Where x is the acid already in solution

Ka = 1.5*10^-2

Substitute in Ka1

Ka1 = [HSO3-][H+] /[H2SO3]

1.5*10^-2 = x*x / (0.030 - x )

Solve for x

x = 0.015 and -0.3

ignore the negative solution since there are no negative concentrations

x = [H+] = [HSO3-] = 0.015

Please remember this number, we will use it at the end when calculating total H+ moles

HINT: we could ignore the second equilibrium since Ka2 is too low compared to the Ka1. That is, the 95% or more of the acid is given by the first ionization

But lets do the other Ka2 equilibrium

Assume, once again

[SO3^-2 ]= [H+] = y NOTE that these H+ ions are from the SECOND ionization and cannot be compared with the first ionization, thats why I'm using x and y to denote difference

[HSO3-] = x-y

[HSO3-] = x will be before the second dissociation, after the dissociation you need to account for the lost acid which is denoted as -y

Ka2 = 6.3*10^-8

Ka2 = [SO3^-2 ][H+]/ [HSO3-]

Substitute all data

6.3*10^-8= y*y / (x-y)

6.3*10^-8 = y² / (0.015-y)

Sovle for y

y = 3.07*10^-5 and -3.07*10^-5

Ignore negative values, no negative concentrations exist

Then

[H+] = y = 3.07*10^-5

Time to add [H+] of first ionization +[H+] of second ionization

that is x+y = 0.015 + 3.07*10^-5 = 0.01503

As you can see, the effect of the second ion is not enough to change the value of x significantively

pH = -log[H+] = -log(0.01503) = 1.82

pH = 1.82

B)

[H+] = 0.01503

[OH-] = 10^-14 / [H+] = (10^-14)/(0.01503) = 6.65*10^-13

[SO3-2] = y = 3.07*10^-5

[HSO3-] = x -y = 0.015 - 3.07*10^-5 = 0.01497

[H2SO3] = 0.030 - x = 0.030 - 0.015 = 0.015