pH, concentration calculation of strong bases | NaOH, KOH

Strong bases dissociate completely in the water. Strong bases may show higher pH values than weak bases (such as aqueous ammonia solution).

In this tutorial,

  1. Calculate pOH, pH when concentration is known
  2. 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.

base dissociation in water

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.

  1. Calculate pOH and then pH
  2. 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.

NaOH dissiciation in water

We have to calculate concentration of OH-(aq) to calculate pH.

strong alkaline dissociation table

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

pH variation against concentration of base


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.

barium hydroxide dissociation

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.

Dissociation of Ba(OH)2

[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.

NaOH solution diluting

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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