Calculate pH of Strong Bases (Alkalis)| NaOH, KOH

Strong bases dissociate completely to hydroxyl ions and related cations in the water. Strong bases show higher pH values than weak bases (such as aqueous ammonia solution). Sodium hydroxide (NaOH), potassium hydroxide (KOH), barium hydroxide are famous strong bases. From those two strong bases, NaOH and KOH has lot of uses in laboratories and chemical industries.

Content

• Definition of strong bases | What are the strong bases or strong alkalis?
  • General equation of strong acid dissociation
  • Examples for strong bases
• pH Calculation steps strong bases solution
  • pH and pOH values
  • Methods to use for calculating pH of a base
• Calculate pH of strong bases such NaOH, KOH
  • Calculate pH of 0.1 moldm^-3 NaOH solution
  • Calculate concentration of KOH when pH 10 at 25⁰C
  • Calculate pH of 0.01 moldm^-3 NaOH solution
  • Calculate concentration and pH of NaOH solution when 4.0g of solid NaOH is dissolved in 250 cm³ distilled water
• pH variation against concentration of base
• Questions

Definition of strong bases | What are the strong bases or strong alkalis?

Strong bases dissociate completely to ions in water. Some of strong bases are Sodium hydroxide (NaOH), Potassium hydroxide (KOH), Barium hydroxide (Ba(OH)₂). In water, these bases dissociate into metal cation and hydroxyl ion (OH^-) ion. As an example, NaOH dissociates completely to Na⁺ ion and OH^- ion.

General equation of strong acid dissociation

Examples for strong bases

Following metal hydroxides dissolve very well in water and give strong alkaline solution.

• Lithium hydroxide (LiOH)
• Sodium hydroxide (NaOH)
• Potassium hydroxide (KOH)
• Rubidium hydroxide (RbOH)
• Barium hydroxide ( Ba(OH)₂ )
• Sodium ethoxide (CH₃CH₂O^-Na⁺)

pH Calculation steps strong bases solution

pH is mainly changed due to change of concentration of the solution. The concentration of the base increases, it results increase of pH.

pH and pOH values

pH is defined for expressing H₃O⁺ concentration and pOH is defined for expressing OH^- concentration in an aqueous solution.

In calculating pH of NaOH solution, first, we have to calculate the pOH value. We have a relationship between pH and pOH as below.

pH + pOH = pKa

pKa is a constant to a certain temperature and pKa = 14 at 25⁰C. So at 25⁰C,

pH + pOH = 14

Methods to use for calculating pH of a base

We can use two methods to calculate pH of basic solution.

1. Calculate pOH of the base and then pH
2. Calculate [H₃O]⁺_(aq), then calculate the pH

Method 1

Calculate pOH of strong base and pH

The formula of pOH,

pOH = -log₁₀[OH^-_(aq)]

calculating pOH is a logarithmic one.

Then use the expression of

pH + pOH = pK_w

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) = K_w

K_w = constant

At 25⁰K_w,
K_w = 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)]

Calculate pH of strong bases such NaOH, KOH, Ba(OH)₂

In this section, we are going to learn how to calculate pH value of strong bases at different concentration and different situations. As well, we will see how to calculate concentration of base when pH value is given.

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.

As the first step, we are going to calculate pOH value and then calculate the pH using the relationship of pH and pOH.

Calculate pOH

• pOH = -log(OH^-_(aq))
• pOH = -log ( 0.1)
• pOH = 1

Apply the 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) = K_w

at 25⁰C, K_w = 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 concentration of KOH when pH 10 at 25⁰C

Here, we are given the pH value of aqueous KOH solution. Now, we have to calculate the concentration of KOH. If we know the pH value, we can calculate pOH from the relation of pH, pOH and pKw

• pH + pOH = 14 at 25⁰C.
• So pOH = 4

From pOH equation, we can calculate OH^- concentration.

• pOH = -log₁₀[OH^-_(aq)]
• 4 = -log₁₀[OH^-_(aq)]
• [OH^-_(aq)] = 0.0001 mo dm^-3

Example 3

Calculate pH of 0.01 mol dm^-3 NaOH solution

Here only changes the concentration of NaOH. As we studied earlier, 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

Apply relation of pH and pOH

• pH + pOH = 14 (at 25⁰C)
• pH + 2 = 14
• pH = 12

Example 3

Calculate pH of 0.1 moldm^-3 of Barium hydroxide ( Ba(OH)₂ )

Barium hydroxide ( Ba(OH)₂ ) is a strong base. Therefore it dissociate completely in water to Ba²⁺_(aq) ions 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 balanced equation to observe stoichiometry ratio.

[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, n_NaOH

n_NaOH = 4 g / 40 gmol^-1
n_NaOH = 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

pH variation against concentration of base

At 25⁰C, pH + pOH = 14

Questions

Is 12.0 mol l^-1 OH^- ion concentration is strong base?

Yes. it is.

If pH of a solution of the strong base NaOH is known, which property of the solution can be calculated?

If pH value is high, we know, pH value will completely depend on the concentration of NaOH. Otherwise we have to consider dissociation of water to calculate OH^- concentration.

• If pH is higher (above 9), we can directly calculate OH^- concentration from pOH = -log₁₀[(OH^-)].
• To find the pOH, use the relationship of pH + pOH = pKa.
• Also you can calculate concentration of [H₃O⁺] from pH value.

Is pH value of different NaOH and KOH solutions equal or not when their concentrations are equal.?

Both NaOH and KOH are strong bases and dissociate completely in the water. So when NaOH and KOH concentrations are equal, after the dissociation, hydroxyl ion concentration is also same. Therefore pH of both solution is equal.

But, basic strength of KOH is much higher than NaOH.

NaOH ph level

pH of dilute NaOH solutions are usually above 10 if concnetration of NaOH is higher than 0.0001 mol dm^-3

How much pH of 0.1 mol dm^-3 KOH

pOH is 1 and pH = 13

0.01 moldm-3 naoh ph

pOH of 0.01 mol dm^-3 NaOH = 2

pH of 0.01 mol dm^-3 NaOH = 12

calculate the ph of a 0.02 mol dm^-3 sodium hydroxide solution in water at 25⁰C given that it contains a 0.02 mol dm^-3 of hydroxide ions

When sodium hydroxide solution is dissolved in water it dissociates completely to sodium ion and hydroxyl ions. According to the balanced equation, concentraion of hydroxyl ions is equal to the concentration of sodium hydroxide.

Therefore, pH of 0.02 mol dm^-3 sodium hydroxide is 1.69

If OH^- has a concentration of 0.00092832 at 25⁰C, what is pH?

OH^- concentration = 0.00092832 mol dm^-3. Substitute this in pOH equation. Then you can use pH + pOH = 14 eqution. At 25⁰C, pH + pOH = 14

which is stronger base KOH or Ni(OH)₂?

Nickel hydroxide (Ni(OH)₂) is a green color precipitate and give very less OH^- amount to water. Therefore, Ni(OH)₂ is a weak basic solution. But, KOH is highly soluble in water and dissociates completely in water to give a strong basic solution.

what is the concentration of Mg(OH)₂ in a solution with a ph of 8.3?

Magnesium hydroxide is a white color precipitate and give very less OH^- amount to water. Therefore, OH^- concentration is low and show pH values are not higher as NaOH, KOH solutions. According to the given pH, we can calculate what is the concentration of Mg(OH)₂.

• pOH + pH = 14
• pOH + 8.3 = 14
• pOH = 5.7

• pOH = -log(OH^-_(aq))
• 5.7 = -log(OH^-_(aq))
• OH^-_(aq) = 2 * 10^-6 mol dm^-3

Assuming all OH^- ions exist in water are given by Mg(OH)₂ (not from water ionization)

Because, we know the OH^- concentration, we can calculate concentration of Mg(OH)₂.

• concentration of Mg(OH)₂ = 2 * 10^-6 mol dm^-3 / 2
• concentration of Mg(OH)₂ = 1 * 10^-6 mol dm^-3 / 2

pH of 0.32 M Mg(OH)₂

Magnesium hydroxide (Mg(OH)₂) is not asoluble hydroxide in water. It forms a weak basic solution. First, you should know that, is it possible to have a concentration like 0.32 M of 0.32 M Mg(OH)₂. You can calculate that from using solubility data of inorganic compounds.

Is KOH a strong base

Yes. KOH is a strong base and it completely dissociates to K⁺ and OH^- ions in the water.

calculate the ph of naoh

Just like KOH, substitute hydroxyl ion concentration given by NaOH to the pOH equation. Then substitute pOH value to the pH + pOH = 14 equation and find pH of NaOH.

How about the pH of aqueous ammonium hydroxide and sodium hydroxide?

Ammonium hydroxide is a weak base give less hydroxyl ion concentration. But sodium hydroxide is a strong acid and completely dissociate to sodium ion and hydroxyl ions.

When hydroxyl ion concentration is high, its basic strength is higher and has a high pH value.

Related tutorials