 # pH of Strong Bases (Alkalis) Calculations Chemistry Tutorial

## Key Concepts

• The pH of an aqueous solution of a strong Arrhenius base can be calculated if we know
(i) the concentration of hydrogen ions in solution
or
(ii) if we know the temperature of the solution and either the pOH of the solution or the concentration of hydroxide ions in solution.
• For a basic (alkaline) solution:
pH = -log10[H+(aq)]1
where [H+(aq)] = concentration of hydrogen ions in solution in mol L-1
• For an aqueous basic (alkaline) solution at 25°C
pH = 14 - pOH
where pOH is the pOH of the solution
• For an aqueous basic (alkaline) solution at 25°C
pH = 14 - (-log10[OH-(aq)])
where [OH-(aq)] is the concentration of hydroxide ions in solution in mol L-1
{Since pOH = -log10[OH-(aq)], and pH = 14 - pOH, pH = 14 - (-log10[OH-(aq)])}

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## Calculating the pH of Strong Arrhenius Bases

Before a strong Arrhenius base is added to water, the water molecules are in equilibrium with hydrogen ions and hydroxide ions:

 water hydrogenions + hydroxide ions H2O H+(aq) + OH-(aq)

Only a very small number of water molecules dissociate into H+(aq) and OH-(aq).

At 25°C, [H+(aq)] = [OH-(aq)] ≈ 10-7 mol L-1 (a very low concentration)
and Kw = [H+(aq)][OH-(aq)] = 10-7 × 10-7 = 10-14

By calculating the logarithm (to base 10) of all the species above, we change the calculation to an addition instead of a multiplication:

Kw = [H+(aq)][OH-(aq)]
log10(Kw) = log10[H+(aq)] + log10[OH-(aq)]

If we then multiply throughout by -1:

-log10(Kw) = -log10[H+(aq)] + -log10[OH-(aq)]
We can see that:
-log10(Kw) = pH + pOH
since pH = -log10[H+(aq)]
and pOH = -log10[OH-(aq)]

For water at 25°C,

Kw = 10-14
log10Kw = log1010-14 = -14
-log10Kw = -log1010-14 = 14
so, 14 = pH + pOH

When a strong Arrhenius base is added to water, the base dissociates completely to form hydrated metal cations and OH-(aq):

 base → hydroxideions + metal cations Group 1 metal hydroxide: MOH → OH-(aq) + M+(aq) Group 2 metal hydroxide: M(OH)2 → 2OH-(aq) + M2+(aq)

Adding the base to the water will disturb the water dissociation equilibrium:

H2O H+(aq) + OH-(aq)
By Le Chatelier's Principle, adding more OH-(aq) to the water will shift the equilibrium position to the left.
The water dissociation equilibrium system responds to the addition of more OH-(aq) by reacting some of the OH-(aq) with some of the H-+(aq) in order to re-establish equilibrium.
So, increasing the concentration of OH-(aq) in the water, reduces the concentration of H+(aq), but, the water dissociation constant does not change2, Kw is still 10-14.
So, Kw = [H+(aq)][OH-(aq)] = 10-14
and -log10Kw = 14 = pH + pOH

[H+(aq)]mol L-1 [OH-(aq)]mol L-1 Kw pH pOH -log(Kw)(pKw) At 25oC 10-7 10-7 10-14 7 7 14 < 10-7 > 10-7 10-14 > 7 < 7 14

We can use the value of Kw or -log10Kw (pKw) and pOH or [OH-(aq)] to calculate pH at a given temperature:

(i) If we know the pOH of the basic (alkaline) solution, we can calculate the pH:
pH + pOH = -log10Kw
So, pH = -log10Kw - pOH
At 25oC : pH = 14 - pOH

(ii) If we know the concentration of hydroxide ions in solution, we can calculate the pH:
Kw = [H+(aq)][OH-(aq)]
By rearranging this equation (formula) we can determine the concentration of hydrogen ions in the aqueous solution:
[H+(aq)] = Kw ÷ [OH-(aq)]
so, pH = -log10[H+(aq)] = -log10(Kw ÷ [OH-(aq)])
Both [H+(aq)] and [OH-(aq)] must be in units of mol L-1 (mol/L or M)

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## Worked Examples

(based on the StoPGoPS approach to problem solving in chemistry.)

Question 1. Calculate the pH of an aqueous solution of sodium hydroxide with a hydrogen (oxidanium or oxonium or hydronium) ion concentration of 1 × 10-13 mol L-1.

1. What have you been asked to do?
Calculate the pH
pH = ?
2. What information (data) have you been given?
Extract the data from the question:
[H+(aq)] = 1 × 10-13 mol L-1
3. What is the relationship between what you know and what you need to find out?
Write the equation (formula) for finding pH:
pH = -log10[H+(aq)]
4. Substitute the values into the equation and solve for pH:
pH = -log10[1 × 10-13] = 13

Question 2. Calculate the pH of an aqueous solution of calcium hydroxide at 25°C if its pOH is 3.4

1. What have you been asked to do?
Calculate the pH
pH = ?
2. What information (data) have you been given?
Extract the data from the question:
pOH = 3.4
temperature = 25°C
so Kw = 1.0 x 10-14 (from Data Sheet)
and -log10Kw = -log1010-14 = 14
therefore:
pH + pOH = 14
Rearrange the equation (formula) to calculate pH:
pH = 14 - pOH
3. Substitute the values into the equation and solve for pOH:
pH = 14 - 3.4 = 10.6

Question 3. 0.40 g of sodium hydroxide is dissolved in enough water to make 500.0 mL of solution at 25°C.
What is the pH of this solution?

(A) Hydroxide ion concentration method:

1. What have you been asked to do?
Calculate the pH
pH = ?
2. What information (data) have you been given?
Extract the data from the question:
mass NaOH(s) = 0.40 g
volume of solution (NaOH(aq)) = 500.0 mL
Convert volume in mL to L by dividing by 1000
volume of solution = 500.0 ÷ 1000 = 0.5000 L
3. What is the relationship between what you know and what you need to find out?
(i) Calculate the moles of sodium hydroxide dissolved in water:
moles = mass ÷ molar mass
mass NaOH = 0.40 g (from the question)
molar mass NaOH = 22.99 + 16.00 + 1.008 = 39.998 g mol-1 (from the Periodic Table)
moles NaOH = 0.40 ÷ 39.998 = 0.010 mol

(ii) Write the equation for the complete dissociation of NaOH in water:
NaOH → Na+(aq) + OH-(aq)

(iii) Determine the stoichiometric ratio (mole ratio) of sodium hydroxide to hydroxide ions:
NaOH : OH-     is 1 : 1 (from the balanced chemical equation)

(iv) Determine the moles of hydroxide ions using the stoichiometric (mole) ratio:
moles OH-(aq) = moles NaOH = 0.010 mol

(v) Calculate the concentration of hydroxide ions in solution in mol L-1 (molarity):
[OH-(aq)] = molarity = moles ÷ volume in litres
[OH-(aq)] = 0.010 ÷ 0.500 = 0.020 mol L-1

(vi) Write the equation (formula) for calculating the concentration of hydrogen ions in water:
Kw = [H+(aq)][OH-(aq)]
So, [H+(aq)] = Kw/[OH-(aq)]

(vii) Substitute in the values for Kw and [OH-(aq)]:
[H+(aq)] = 1.0 ×l 10-14/0.020 = 5.0 × 10-13 mol L-1

(viii) Write the equation (formula) for calculating pH:
pH = -log10[H+(aq)]

4. Substitute in the value for [H+(aq)] and solve:
pH = -log10[5.0 × 10-13] = 12.3

(B) pOH method:

1. What have you been asked to do?
Calculate the pH
pH = ?
2. What information (data) have you been given?
Extract the data from the question:
mass NaOH(s) = 0.40 g
volume of solution (NaOH(aq)) = 500.0 mL = 500.0 ÷ 1000 = 0.5000 L
3. What is the relationship between what you know and what you need to find out?
(i) Calculate the moles of sodium hydroxide dissolved in water:
moles = mass ÷ molar mass
mass NaOH = 0.40 g (from the question)
molar mass NaOH = 22.99 + 16.00 + 1.008 = 39.998 g mol-1 (from the Periodic Table)
moles NaOH = 0.40 ÷ 39.998 = 0.010 mol

Write the equation for the complete dissociation of NaOH in water:
NaOH → Na+(aq) + OH-(aq)

(ii) Determine the stoichiometric ratio (mole ratio) of sodium hydroxide to hydroxide ions:
NaOH : OH-     is 1 : 1 (from the balanced chemical equation)

(iii) Determine the moles of hydroxide ions using the stoichiometric (mole) ratio:
moles OH-(aq) = moles NaOH = 0.010 mol

(iv) Calculate the concentration of hydroxide ions in solution in mol L-1 (molarity):
[OH-(aq)] = molarity = moles ÷ volume in litres
[OH-(aq)] = 0.010 ÷ 0.500 = 0.020 mol L-1

(v) Write the equation (formula) for calculating pOH:
pOH = -log10[OH-(aq)]

(vi) Substitute in the value for [OH-(aq)] and solve:
pOH = -log10[0.020] = 1.7

(vii) Write the equation (formula) for calculating pH from Kw:
-log10Kw = pH + pOH
(or pKw = pH + pOH)

4. Substitute in the values for Kw and pOH and solve for pH:
-log10[1.0 × 10-14] = pH + 1.7
14 = pH + 1.7
14 - 1.7 = pH = 12.3

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1. For simplicity we will use the symbol H+(aq) to refer to a hydrated hydrogen ion, which is equivalent to the oxidanium (oxonium or hydronium) ion, H3O+.

2. This assumes that the temperature of the water remains constant.
If the temperature changes, the value of Kw changes.