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pOH of Strong Bases (Alkalis) Calculations Tutorial

Key Concepts

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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 LiOHOH-(aq) + Li+(aq)
sodium hydroxide NaOHOH-(aq) + Na+(aq)
potassium hydroxide KOHOH-(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
MOHOH-(aq) + M+(aq)   M(OH)22OH-(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
MOHOH-(aq) + M+(aq)   M(OH)22OH-(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-]

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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
  5. 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.
  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
  5. 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.

  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
  5. 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.

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

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