What is the necessary volume and molarity to dilute sulfuric acid to have a pH of 4?

Question

For my G4 experiment, I was tasked to explore a topic that covered both chemistry in the environment and I immediately chose acid rain. When sulfur dioxide reacts with oxygen and water in the gaseous state, it creates sulfuric acid (acid rain) and I wanted to explore the effect that different pH of sulfuric acid would have on plants, fruits, etc. I have successfully created 2 PH concentrations of sulfuric acid at 2.4 pH and 6 pH but am struggling to find a correct start/ending molarity to make 1 liter of sulfuric acid at a pH of 4?

With M1V1=M2V2

I only know that the ending volume I want is 1 Liter. My only problem is that sulfuric acid is diprotic and I have no yet learned nor do I understand how dilution works to make it have a pH of 4.

Answer 1

When H2SO4 is placed in water, the following dissociation takes place:

H2SO4 ===> H⁺ + HSO4^-  Ka1 = large

HSO4- ===> H⁺ + SO4^2-  Ka2 = 1.2x10^-2

Note that Ka2 is relatively insignificant compared to Ka1 and in most cases it is NOT considered when calculating pH. 

 

For pH = 4, the [H+] = 1x10^-4 M

Considering ONLY Ka1, [H2SO4] = 1x10^-4 and pH = 4

So for practical purposes, you can prepare a 1x10-4 M H2SO4 solution and the pH should be very close to 4, assuming neutral H2O, which isn't always the case.  If you dilute the pH 6 solution that you have 1:100, i.e. 10 ml to 1 liter, the pH should be close to 4.