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 going to learn followings.
We can dilute any solution by adding strong acid to the distilled water. In strong acidic solution, when the acidic solution is diluted, H3O+ concentration decreases and pH increases.
But keep in mind whether diluting of weak acid can be done by using distilled water too.
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
First, we look some examples which will help to get an idea about concentrations after diluting.
Question
There is a 0.01 mol dm-3 hydrochloric acid, 50 cm3 solution in a beaker. If another 50 cm3 distilled water is added to the hydrochloric acid solution, what is the new concentration of acid?
Answer
There are two solutions. In one solution, contains aqueous HCl while other solution contains only distilled water. In HCl solution, there is some amount (mol) of HCl. When you add distill water to that HCl solution, amount of HCl is unchanged. Volume of the HCl solution increases due to addition of distill water. Though amount of HCl is unchanged, concentration of HCl is changed because volume of the solution increases.
First calculate the amount of hydrochloric acid (HCl) .
Amount of HCl = 0.01 mol dm-3 * 50 cm3
Amount of HCl = 0.0005 mol
After adiing 50 cm3 of distilled water total volume of the solution is 100 cm3.
Concentration of final HCl = 0.0005 mol / ( 100 / 1000 ) dm3
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.
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) + H2O(l) → Cl- (aq) + H3O+(aq)
Due to completeness dissociation,
[ HCl(aq)] = [H3O+(aq) ]
Assumption: The H3O+ receive from dissociation of water is negligible when it compares with H3O+ 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 10cm3 of 0.1 mol dm-3 solution of H2SO4.
You have to do
1. dilute the initial solution by 10 times
2. dilute the initial solution by 100 times
Find the concentration of H2SO4 and pH of every solution.
you can assume that both dissociation of H2SO4 is complete.
Sulfuric acid is strong acid and it dissociates completely in the water and gives two H3O+
(or H+) ions. Therefore concentration of H3O+ is twice as concentration of
H2SO4.
[H3O+(aq)] = 2*[H2SO4(aq)]
In this example, at initial state,
[H2SO4(aq)] = 0.1 mol dm-3
[H3O +(aq)] = 0.2 mol dm-3
Now we can find the pH of initial solution,
pH = -log10[H3O+(aq)]
pH = -log10[0.2]
pH = 0.699
Concentration of H2SO4(aq) is reduced by 10 times. With that
H3O+(aq) concentration is also reduced by 10 times. Then,
[H2SO4(aq)] = 0.01 mol dm-3
[H3O +(aq)] = 0.02 mol dm-3
Now we can find the pH of solution,
pH = -log10[H3O+(aq)]
pH = -log10[0.02]
pH = 1.699
Concentration of H2SO4(aq) is reduced by 100 times. With that
H3O+(aq) concentration is also reduced by 100 times. Then,
[H2SO4(aq)] = 0.001 mol dm-3
[H3O +(aq)] = 0.002 mol dm-3
Now we can find the pH of solution,
pH = -log10[H3O+(aq)]
pH = -log10[0.002]
pH = 2.699
Have Questions?
There are several steps to do this calculation.
According to the concentration, pH vary. But, usually it exists between 1 - 4 most times. When acid strength is high, pH value become low.
Yes. You can. Let's considera strong acid solution. In that acid, there is a some amount of acid (mol) and volume. When we add water. volume of the solution increases. When volume of the solution increases, concentration of H+ ion is decreased. When H+ ion concentration is changed, pH is also changed.
Few calculations are required. Also you need to know what is the required volume (V1) of the diluted solution.
When you add pure water, acid is diluted. If you want to dilute the acid to a certain concentration or pH, you have to do a calculation which calculates required amount of distill water.
When concentrated acid is diluted, hydronium ion concentration decreases which results a higher pH value. 0.1 M strong acid is diluted upto 0.01 M. Then calculate pH for two cases and compare two results. You will see pH of 0.01 M is higher than 0.1 M solution.
Add strong acid to the distiled water. Never add water to the strong acid solution. Adding distilled water to the strong acid solution, will produce large heat which can cause injuries to people.
Both strong acid and weak acid can be dilluted to decrease the concentration. Add distilled water to the stronger acid to dilute.
If an acid dissociate completely in water to hydronium ion and anion, we call it is a strong acid. Strong acid may be concentrated or dilute.
If acid concentration is low, acid is dilute.
When water is added, volume of solution increases. According to the C = n/V equation, concentration should be decreased because n keeps unchanged and V is increased, Therefore concentration of acid decreases when diluted with water.
For diluting, distilled water should be used. Normal water contain salts and different constituents. Therefore diliting from a normal water may contaminate the strong acid solution
This is different type of question. I can give an answer, but it may be not correct 100%. If we find a method which can remove water from dilute acid solution, we can make a strong acid. But acidic components may be not removed or removing sgould be very low value.
We can evaporate water. But acids such as HCl are also evaporated easily when concentration of acid increases.
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