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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:
| | Arrhenius
base | → | hydroxide
ions | + | 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 base
in 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 base
in 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 base
in 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 base
in 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 base
in 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-]
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).
- What have you been asked to do?
Calculate the pOH
pOH = ?
- What information (data) have you been given?
Extract the data from the question:
[KOH(aq)] = 0.2 mol L-1
- 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-]
- Substitute in the value for [OH-] and solve:
pOH = -log10[OH-]
= -log10[0.2]
= 0.7
- Is your answer plausible?
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.
- 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.
- What have you been asked to do?
Calculate the pOH
pOH = ?
- 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
- 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-]
- Substitute in the value for [OH-] and solve:
pOH = -log10[OH-]
= -log10[0.020]
= 1.7
- Is your answer plausible?
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.
- 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.
- What have you been asked to do?
Calculate the pOH
pOH = ?
- 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
- 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)]
- Substitute in the value for [OH-(aq)] and solve:
pOH = -log10[OH-(aq)]
= -log10[0.012]
= 1.9
- Is your answer plausible?
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.
- State your solution to the problem:
pOH = 1.9
