 # pOH of Strong Bases (Alkalis) Calculations Tutorial

## Key Concepts

• A strong Arrhenius base is one that dissociates completely in water to form hydroxide ions:

 base → hydroxide ions + cations Hydroxides of Group 1 metals: MOH → OH-(aq) + M+(aq) Hydroxides of Group 2 metals: M(OH)2 → 2OH-(aq) + M2+(aq)

• The concentration of hydroxide ions in solution is dependent on the concentration of the original base:

Group 1 metal hydroxides
basehydroxide
ions
+ cation
MOH OH-(aq) + M+(aq)
n moles
of MOH
= n moles
of OH-(aq)
=n moles
of M+(aq)
[OH-(aq)] = [MOH]
Group 2 metal hydroxides
basehydroxide
ions
+ cation
M(OH)2 2OH-(aq) + M2+(aq)
n moles
of M(OH)2
2 x n moles
of OH-(aq)
n moles
of M2+(aq)
[OH-(aq)] = 2 x [M(OH)2]

Square brackets, [ ], are Chemist's short-hand for concentration in mol L-1 (molarity or molar concentration).
• pOH is a measure of the hydroxide ion (OH-) concentration in a solution.
• pOH of a solution is calculated using the formula (equation):
pOH = -log10[OH-]
where [OH-] is the concentration of hydroxide ions in mol L-1 (mol/L or M).

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## Calculating the pOH of Strong Arrhenius Bases

Step 1. Write the equation for the complete dissociation of the strong Arrhenius base in water:

 Arrheniusbase → hydroxideions + cation lithium hydroxide LiOH → OH-(aq) + Li+(aq) sodium hydroxide NaOH → OH-(aq) + Na+(aq) potassium hydroxide KOH → OH-(aq) + K+(aq) calcium hydroxide Ca(OH)2 → 2OH-(aq) + Ca2+(aq) barium hydroxide Ba(OH)2 → 2OH-(aq) + Ba2+(aq)
Common Strong Arrhenius Bases
Group 1 metal hydroxides:

LiOH, NaOH, KOH, RbOH, CsOH

Group 2 metal hydroxides"

Ca(OH)2, Sr(OH)2, Ba(OH)2

Step 2. Use the concentration of the undissociated base to determine the concentration of hydroxide ions in the aqueous solution:

 Group 1 metal hydroxide Group 2 metal hydroxide MOH → OH-(aq) + M+(aq) M(OH)2 → 2OH-(aq) + M2+(aq) for 1 mole of base : 1 mole MOH → 1 mole OH-(aq) + 1 mole M+(aq) 1 mole M(OH)2 → 2 mole OH-(aq) + 1 mole M2+(aq) for 0.1 mole of base : 0.1 mole MOH → 0.1 mole OH-(aq) + 0.1 mole M+(aq) 0.1 mole M(OH)2 → 0.2 mole OH-(aq) + 0.1 mole M2+(aq) for 0.5 mole of base : 0.5 mole MOH → 0.5 mole OH-(aq) + 0.5 mole M+(aq) 0.5 mole M(OH)2 → 1.0 mole OH-(aq) + 0.5 mole M2+(aq) so for n mole of base : n mole MOH → n mole OH-(aq) + n mole M+(aq) n mole M(OH)2 → 2 x n mole OH-(aq) + n mole M2+(aq)

Concentration in mol L-1 (molarity or molar concentration) is calculated by dividing moles by volume in litres:
molarity = moles ÷ volume
The volume of the solution is the same for both the undissociated base, MOH or M(OH)2, and for the hydroxide ions, OH-, it produces.

 Group 1 metal hydroxide Group 2 metal hydroxide MOH → OH-(aq) + M+(aq) M(OH)2 → 2OH-(aq) + M2+(aq) for n mole of basein 1 L of solution: [MOH]=n/1 → [OH-(aq)]=n/1 + [M+(aq)]=n/1 [M(OH)2]=n/1 → [OH-(aq)]=2n/1 =2n + [M2+(aq)]=n/1 for n mole of basein 2 L of solution: [MOH]=n/2 → [OH-(aq)]=n/2 + [M+(aq)]=n/2 [M(OH)2]=n/2 → [OH-(aq)]=2n/2 =n + [M2+(aq)]=n/2 for n mole of basein 0.4 L of solution: [MOH]=n/0.4 → [OH-(aq)]=n/0.4 + [M+(aq)]=n/0.4 [M(OH)2]=n/0.4 → [OH-(aq)]=2n/0.4 + [M2+(aq)]=n/0.4 for n mole of basein 1.3 L of solution: [MOH]=n/1.3 → [OH-(aq)]=n/1.3 + [M+(aq)]=n/1.3 [M(OH)2]=n/1.3 → [OH-(aq)]=2n/1.3 + [M2+(aq)]=n/1.3 for n mole of basein V L of solution: [MOH]=n/V → [OH-(aq)]=n/V + [M+(aq)]=n/V [M(OH)2]=n/V → [OH-(aq)]=2n/V + [M2+(aq)]=n/V

The concentration of the hydroxide ions produced by a strong Arrhenius base:
 Group 1 metal hydroxide: [OH-(aq)] = [MOH] Group 2 metal hydroxide: [OH-(aq)] = 2 × [M(OH)2]

Step 3. Use the concentration of hydroxide ions in mol L-1, [OH-], to calculate the pOH of the solution:

pOH = -log10[OH-]

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## Worked Examples

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

Question 1. Find the pOH of 0.2 mol L-1 KOH(aq).

1. What have you been asked to do?
Calculate the pOH
pOH = ?
2. What information (data) have you been given?
Extract the data from the question:
[KOH(aq)] = 0.2 mol L-1
3. What is the relationship between what you know and what you need to find out?
Write the balanced chemical equation for the dissociation of KOH in water:
KOH → K+(aq) + OH-(aq)

Use the balanced chemical equation to calculate the concentration of hydroxide ions in solution in mol L-1:
[Group 1 metal hydroxide] = [OH-]
[KOH(aq)] = [OH-] = 0.2 mol L-1

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

4. Substitute in the value for [OH-] and solve:
pOH = -log10[OH-]
= -log10[0.2]
= 0.7
Round off the concentration of KOH given from 0.2 to 0.1 (= 10-1)
pOH = -log10[10-1] = 1
Since our calculated answer is close to this approximation, we are confident are answer is correct.
6. State your solution to the problem:
pOH = 0.7

Question 2. 0.00100 mol of barium hydroxide is dissolved in water to make 0.10 L of aqueous solution.
Calculate the pOH of the aqueous barium hydroxide solution.

1. What have you been asked to do?
Calculate the pOH
pOH = ?
2. What information (data) have you been given?
Extract the data from the question:
moles of Ba(OH)2(aq) = n(Ba(OH)2(aq)) = 0.0010 mol
volume of solution = V = 0.10 L
3. What is the relationship between what you know and what you need to find out?
Write the balanced chemical equation for the dissociation of Ba(OH)2 in water:
Ba(OH)2 → 2OH-(aq) + Ba2+(aq)

Use the balanced chemical equation to calculate the moles of hydroxide ions in solution:
stoichiometric ratio (mole ratio) of Ba(OH)2 : OH- is 1 : 2
0.0010 moles Ba(OH)2 produces 2 × 0.0010 moles OH-
moles of OH-(aq) = n(OH-(aq)) = 0.0020 moles

Calculate the concentration of hydroxide ions in solution in mol L-1:
[OH-(aq)] = moles OH-(aq) ÷ volume of solution in L
[OH-(aq)] = 0.0020 ÷ 0.10 = 0.020 mol L-1

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

4. Substitute in the value for [OH-] and solve:
pOH = -log10[OH-]
= -log10[0.020]
= 1.7
work backwards using calculated pOH value to calculate [OH-]:
pOH = -log10[OH-]
so [OH-] = 10-pOH = 10-1.7 = 0.02 mol L-1

Calculate [Ba(OH)2]:
[Ba(OH)2] = ½ × [OH-] = ½ × 0.02 = 0.01 mol L-1

Calculate moles Ba(OH)2 in 0.10 L of solution:
moles = molarity × volume = 0.01 × 0.10 = 0.001 mol

This is the same as the moles Ba(OH)2 given in the question so we are confident our answer is correct.

6. State your solution to the problem:
pOH = 1.7

Question 3. 0.12 g of sodium hydroxide is dissolved in enough water to make 0.25 L of solution.
Calculate the pOH of the sodium hydroxide solution.

1. What have you been asked to do?
Calculate the pOH
pOH = ?
2. What information (data) have you been given?
Extract the data from the question:
mass of NaOH(aq) = 0.12 g
volume of solution = V = 0.25 L
3. What is the relationship between what you know and what you need to find out?
Calculate the moles of NaOH:
moles NaOH = mass NaOH in grams ÷ molar mass of NaOH in g mol -1
molar mass NaOH = 22.99 + 16.00 + 1.008 = 39.998 g mol -1 (use Periodic Table to find molar mass of each atom)
moles NaOH = 0.12 g ÷ 39.998 g mol -1
n(NaOH) = 3.00 × 10-3 mol

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

Use the balanced chemical equation to calculate the moles of hydroxide ions in solution:
stoichiometric ratio (mole ratio) of NaOH : OH- is 1 : 1
3.00 × 10-3 moles NaOH produces 3.00 × 10-3 moles OH-

Calculate the concentration of hydroxide ions in solution in mol L-1:
[OH-(aq)] = moles OH-(aq) ÷ volume of solution in L
[OH-(aq)] = 3.00 × 10-3 ÷ 0.25
= 0.012 mol L-1

pOH = -log10[OH-(aq)]

4. Substitute in the value for [OH-(aq)] and solve:
pOH = -log10[OH-(aq)]
= -log10[0.012]
= 1.9
Round up pOH from 1.9 to 2 then use this to calculate the mass of NaOH needed to make 0.25 L of solution.
pOH = -log10[OH-(aq)]
so [OH-(aq)] = 10-pOH = 10-2 mol L-1

Calculate [NaOH(aq)]:
[NaOH(aq)] = [OH-(aq)] = 10-2 mol L-1

Calculate moles NaOH in 0.25 L solution:
moles(NaOH) = molarity × volume = 10-2 × 0.25 = 2.5 × 10-3 mol

Calculate mass of NaOH in 2.5 × 10-3 mol
mass = moles × molar mass = 2.5 × 10-3 × (23 + 16 + 1) = 0.1 g

Since this approximate mass is about the same as that given in the question, we are confident our answer is correct.

6. State your solution to the problem:
pOH = 1.9

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