Strong acids dissociate completely to H+ ions and anion. When acid is diluted, concentration decreases and there is nice relationship between pH and diluting times such as diluted by 10 times, 100 times and more.

We can dilute any solution by adding distilled water. In acidic solution, when the acidic solution is diluted, H_{3}O^{+}
concentration decreases and pH increases

You should have an understand about **how to find pH of strong acids**.
If you don't have please refer following tutorial.

Find pH of strong acids

There are 3 bottles which contain aqueous hydrochloric (HCl) acid solutions. These hydrochloric acids' concentrations in
mol dm^{-} are 0.1, 0.01, 0.001. Calculate pH of those solutions.

Hydrochloric acid is a strong acid. Therefore it dissociate completely in the water.

HCl_{(aq)} + H_{2}O_{(l)} → Cl^{-} _{(aq)} +
H_{3}O^{+}_{(aq)}

Due to completeness dissociation,

[ HCl_{(aq)}] = [H_{3}O^{+}_{(aq)} ]

Assumption: The H_{3}O^{+} receive from dissociation of water is negligible when it compares with
H_{3}O^{+} receive from HCl.

We can see a pattern of pH values of solutions. pH variation against concentration of strong acids is figured below. you can see diluting 10 times will increase pH value by 1.

When weak acid is diluted by 10 times, concentration of it is decreased by 10 times.

When strong acid is diluted by 10 times, pH value is increased by 1

When strong acid is diluted by 100 times, pH value is increased by 2

When strong acid is diluted by 1000 times, pH value is increased by 3

Example problem:

You are provided 10cm^{3} of 0.1 mol dm^{-3} solution of H_{2}SO_{4}.
You have to do

1. dilute the initial solution by 10 times

2. dilute the initial solution by 100 times

Find the concentration of H_{2}SO_{4} and pH of every solution.

**you can assume that both dissociation of H _{2}SO_{4} is complete.**

Sulfuric acid is strong acid and it dissociates completely in the water and gives

[H

In this example, at initial state,

[H_{2}SO_{4(aq)}] = 0.1 mol dm^{-3}

[H_{3}O ^{+}_{(aq)}] = 0.2 mol dm^{-3}

Now we can find the pH of initial solution,

pH = -log_{10}[H_{3}O^{+}_{(aq)}]

pH = -log_{10}[0.2]

pH = 0.699

Concentration of H_{2}SO_{4(aq)} is reduced by 10 times. With that
H_{3}O^{+}_{(aq)} concentration is also reduced by 10 times. Then,

[H_{2}SO_{4(aq)}] = 0.01 mol dm^{-3}

[H_{3}O ^{+}_{(aq)}] = 0.02 mol dm^{-3}

Now we can find the pH of solution,

pH = -log_{10}[H_{3}O^{+}_{(aq)}]

pH = -log_{10}[0.02]

pH = 1.699

Concentration of H_{2}SO_{4(aq)} is reduced by 100 times. With that
H_{3}O^{+}_{(aq)} concentration is also reduced by 100 times. Then,

[H_{2}SO_{4(aq)}] = 0.001 mol dm^{-3}

[H_{3}O ^{+}_{(aq)}] = 0.002 mol dm^{-3}

Now we can find the pH of solution,

pH = -log_{10}[H_{3}O^{+}_{(aq)}]

pH = -log_{10}[0.002]

pH = 2.699

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