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pH of Strong Bases (Alkalis) Calculations Tutorial

Key Concepts

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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 hydrogen
ions
+ 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 hydroxide
ions
+ 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 At 25oC pH pOH -log(Kw)
(pKw)
10-7 10-7 10-14 pure water
(before base added)
7 7 14
< 10-7 > 10-7 10-14 aqueous solution
(after base added)
> 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 (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 hydronium ion, H3O+.

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