Question

Determine the pH of a 0.034 solution of H2SO4. The dissociation occurs in two steps. Ka1...

Determine the pH of a 0.034 solution of H2SO4. The dissociation occurs in two steps. Ka1 is extremely large; Ka2 is 1.2 * 10^-2
Please show work thanks

Homework Answers

Answer #1

Ka1 is large so the first proton is fully dissociated

This gives [H+] of 3.4 x10-2 = 0.034

The second proton is weakly dissociated so

Ka2 = [H+][SO42-]/[HSO4-)] = 1.2x10-2 = 0.012

We know that the total concentration of HSO4- present is the same as the initial concetration of H2SO4 because the first dissociation is complete.

Let the amount of dissociation of HSO4- be x.

After dissociation [HSO4- ]= (0.0032 - x) and [H+] = [SO42-] = x

But from Ka1 we already have [H+] of 0.0032 so total [H+] = 0.0032 + x

From Ka2:

0.012 = (0.0032 + x) x/(0.0032 - x)

0.012 (0.0032 - x) = (0.0032 + x) x

0.0000384 - 0.012x = 0.0032x + x2

x2 + 0.0152x - 0.0000384 = 0

Solve this quadratic for x using the quadratic formula.

on calculation we get, x = 0.0022

This is the concentration of H+ from the second dissociation

Total [H+] = 0.0032 + 0.0022 = 0.0054

pH = - log(.0054) = 2.27

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