Strong acids dissociate completely to hydrogen ion and anion. When acid is diluted, concentration decreases and there is clear relationship between pH and number of diluting times. In this tutorial, we are goint to obtain a relationship between diluting factor and pH.

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

- No change in amount (mol) of acid
- But due to change of volume of solution (increase), concentration is changed (reduced).

You should have an understanding about **how to find pH of strong acids**.
If you don't have please refer following tutorial for better undersatnding of this lesson.

Find pH of strong acids

- Determine the concentration and volume of required diluted acid solution. Then you can know the exact amount of requirement of acid.
- Calculate the required volume of strong acid (concentrated) which can give the amount in dilute solution. Use C
_{1}V_{1}= C_{2}V_{2}relation. - Then separate the required volume from high concentrated solution. Add that solution to water slowly for diluting until required volume of dilute acid is completed.

- Hydrochloric acid (HCl)
- Nitric acid (HNO
_{3}) - Sulfuric acid (H
_{2}SO_{4}) - Perchloric acid (HClO
_{4})

First, we look some examples which will help to get an idea about concentrations after diluting.

There is a 0.01 mol dm^{-3} hydrochloric acid 50 cm^{3} solution. If another 50 cm^{3} distilled water is added to the hydrochloric acid solution, what is the concentration of acid?

Answer

First calculate the amount of hydrochloric acid.

Amount of HCl = 0.01 mol dm^{-3} * 50 cm^{3}

Amount of HCl = 0.0005 mol

After adiing 50 cm^{3} of distilled water total volume of the solution is 100 cm^{3}.

Concentration of final HCl = 0.0005 mol / ( 100 / 1000 ) dm^{3}

Concentration of final HCl = 0.005 mol dm^{-3}

Concentration is reduced by a half when we add same volume (initial HCl solution volume) of distilled water. So concentration is dilluted by two times.

H^{+} concentration also decreases by two times in this example.

- Dilute by 2 times, concentration will decrease by two times.
- dilute by 5 times, concentration will decrease by five times.
- dilute by 10 times, concentration will decrease by 10 times.

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

Questions

Both strong acid and weak acid can be dilluted to decrease the concentration. Add distilled water to the stronger acid to dilute.

For diluting, distilled water should be used. Normal water contain salt and different constituents.

In the laboratories, strong acids have higher concentrations. But in our cases, we do not need that kind of higher concentrations. So we do, take little bit of high concentrated acid and add distilled water to dilute it until our reuired concentration.

Related tutorials

browse topics