pH, concentration calculation of strong bases | NaOH, KOH
In this tutorial,
- Calculate pOH, pH when concentration is known
- Calculate concentration when pH is known
What are the strong bases or alkalis?
Strong bases dissociate completely in water. Some of strong bases are NaOH, KOH, Sr(OH)2.In water, these bases dissociate into metal ion and OH- ion. As an example, NaOH dissociate into Na+ and OH- ions.
Calculate pH of strong bases
pH is mainly changed due to change of concentration of the solution. The concentration of the base increases, it results increase of pH.
First we have to calculate the pOH.
We can use two methods to calculate pH of basic solution.
- Calculate pOH and then pH
- Calculate [H3O]+(aq), then calculate the pH
Method 1
Calculate pOH and then pH
The formula of pOH,
pOH = -log10[OH-(aq)]
calculating pOH is a logarithmic one.
Then use the expression of
pH + pOH = pKw
to calculate pH. The sum of pH + pOH become 14 when room temperature is 250C. In other temperatures, we have to give the related value of that temperature.
Method 2
Calculate [H3O]+(aq) and then pH
There is a relation between [H3O]+(aq) and [OH]-(aq) in aqueous solution.
Relationship of [H3O]+(aq) and [OH]-(aq)
[H3O]+(aq) * [OH]-(aq) = Kw
Kw = constant
At 250Kw,
Kw = 1.0 * 10-14mol2 dm-6
When we know [OH]-(aq), using this equation, we can easily calculate the [H3O]+(aq).
Then, we can use pH equation, to calculate pH.
pH = -log10[H3O+(aq)]
NOTE
pKw = -log10[Ka]
Calculate pH of strong base examples
Example 1
Calculate pH of 0.1 moldm-3 NaOH solution
The room temperature is 250C.
Note : M = mol dm-3
We solve this example according to method 1.
NaOH dissociate completely into Na+(aq) & OH-(aq) ions in water.
We have to calculate concentration of OH-(aq) to calculate pH.
First we calculate pOH and then calculate the pH using the relationship of pH and pOH
Calculate pOH
pOH = -log(OH-(aq))
pOH = -log ( 0.1)
pOH = 1
relation of pH and pOH
pH + pOH = 14 (at 250C)
pH + 1 = 14
pH = 13
Let's solve example 1 according to method 2.
We know the concentration of OH- ions
[OH-(aq)] = 0.1 moldm-3
Use the relation of,
[H3O]+(aq) * [OH]-(aq) = Kw
at 250C, Kw = 1.0 * 10-14mol2 dm-6
[H3O]+(aq) * 0.1 = 1.0 * 10-14
[H3O]+(aq) = 1.0 * 10-13
Now, we know the concentration of [H3O]+(aq), therefore we can use pH equation,
pH = -log10[H3O+(aq)]
pH = -log10(1.0 * 10-13)
pH = 13
Example 2
Calculate pH of 0.01 moldm-3 NaOH solution
Here only changes the concentration of NaOH. We studied, NaOH completely dissociate in the water. Therefore,
[OH]-(aq) = 0.01 moldm-3
Then substitute [OH]-(aq) to the pOH equation.
pOH = -log(OH-(aq))
pOH = -log ( 0.01)
pOH = 2
relation of pH and pOH
pH + pOH = 14 (at 250C)
pH + 2 = 14
pH = 12
pH variation against concentration of base
When OH- concentration of aqueous solution is reduced by 10 times, pH is increased by a 1.
At 250C, pH + pOH = 14
Example 3
Calculate pH of 0.1 moldm-3 of Ba(OH)2
Ba(OH)2 is a strong base. Therefore it dissociate completely in water to Ba2+(aq) and OH- ions.
Dissociation of Ba(OH)2
When one Ba(OH)2 molecule dissociate, one Ba2+ ion and two OH- ions are given. See the stoichiometry ratio.
2.jpg)
[OH-(aq)] = 0.2 moldm-3
Then substitute [OH]-(aq) to the pOH equation.
pOH = -log(OH-(aq))
pOH = -log ( 0.2)
pOH = 0.7
relation of pH and pOH
pH + pOH = 14 (at 250C)
pH + 0.7 = 14
pH = 13.3
Example 4
Calculate concentration and pH of NaOH solution
4g of solid NaOH are dissolved in water and then diluted the mixture upto 250cm3. Calculate the concentration of OH- ions and pH of solution.
First, number of moles(n) of dissolved NaOH should be calculated.
Molecular weight of NaOH
NaOH = 23 + 16 + 1
NaOH = 40 g mol-1
number of moles of dissolved NaOH, nNaOH
nNaOH = 4 g / 40 gmol-1
nNaOH = 0.1 mol
Calculate the concentration now
[NaOH(aq)] = 0.1mol / 0.25 dm3
[NaOH(aq)] = 0.4 mol dm-3
[OH-(aq)] = 0.4 mol dm-3
pOH = -log(OH-(aq))
pOH = -log ( 0.4)
pOH = 0.4
pH = 13.6
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