Hydroxide Ion Concentration of Strong Bases Chemistry Tutorial

pOH can be used to calculate the concentration of hydroxide ions in solutions of strong Arrhenius bases:
[OH^-] = 10^-pOH
where [OH^-] is the concentration of hydroxide ions in mol L^-1 (mol/L or M).

A strong base is one that fully dissociates in water.
An aqueous solution of a strong base contains hydrated metal ions and hydroxide ions.
An aqueous solution of a strong base does not contain 'molecules' of the undissociated base¹.

For the hydroxides of Group 1 (IA or Alkali) metals, MOH, the concentration of the metal ions in solution, [M⁺_(aq)], is the same as the concentration of hydroxide ions in aquoeus solution [OH^-_(aq)]:
MOH → M⁺_(aq) + OH^-_(aq)
n moles/litre M⁺_(aq) = n moles/litre OH^-_(aq)
[M⁺_(aq)] = [OH^-_(aq)]

For the hydroxides of Group 2 (IIA or Alkaline Earth) metals, M(OH)₂, the concentration of metal ions in solution, [M²⁺_(aq)], is half the concentration of hydroxide ions in aquoeus solution [OH^-_(aq)]:
M(OH)₂ → M²⁺_(aq) + 2OH^-_(aq)
n moles/litre M²⁺_(aq) = 2 × n moles/litre OH^-_(aq)
[OH^-_(aq)] = 2 × [M²⁺_(aq)]
or
½ × n moles/litre M²⁺_(aq) = n moles/litre OH^-_(aq)
[M²⁺_(aq)] = ½ × [OH^-_(aq)]

Please do not block ads on this website.
No ads = no money for us = no free stuff for you!

Calculating the Concentration of Hydroxide Ions in Dilute Aqueous Solutions of Strong Bases

Step 1. Write the equation for finding [OH^-_(aq)]:

[OH^-_(aq)] = 10^-pOH

Step 2. Substitute in the value for pOH and solve to give the concentration of OH^-_(aq) in mol L^-1 (mol/L or M).

Do you know this?

Join AUS-e-TUTE!

Play the game now!

Calculating the Concentration of Metal Ions in Dilute Aqueous Solutions of Strong Bases

Step 1. Write the equation for finding [OH^-_(aq)]:

[OH^-_(aq)] = 10^-pOH

Step 2. Substitute in the value for pOH and solve to give the concentration of OH^-_(aq) in mol L^-1 (mol/L or M).

Step 3. Write the balanced chemical equation for the dissociation of the strong base in water:

	Group 1 hydroxide	Group 2 hydroxide
Chemical Equation	MOH → M⁺_(aq) + OH^-_(aq)	M(OH)₂ → M²⁺_(aq) + 2OH^-_(aq)

Step 4: Determine the stoichiometric ratio (mole ratio) of hydroxide ions to metal ions in solution:

	Group 1 hydroxide	Group 2 hydroxide
Chemical Equation	MOH → M⁺_(aq) + OH^-_(aq)	M(OH)₂ → M²⁺_(aq) + 2OH^-_(aq)
moles metal ions : moles OH^-	1 : 1	1 : 2

Step 5: Assuming that the volume of the aqueous solution does not change, use the stoichiometric ratio (mole ratio) to determine the concentration of metal ions in solution:

	Group 1 hydroxide	Group 2 hydroxide
Chemical Equation	MOH → M⁺_(aq) + OH^-_(aq)	M(OH)₂ → M²⁺_(aq) + 2OH^-_(aq)
moles metal ions : moles OH^-	1 : 1	1 : 2 ½ : 1
[metal ions_(aq)] =	1 × [OH^-] mol L^-1	½ × [OH^-] mol L^-1

Step 6: Substitute in the value for [OH^-] calculated in Step 2, and solve.

Do you understand this?

Join AUS-e-TUTE!

Take the test now!

Worked Examples

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

Question 1. Calculate the concentration of hydroxide ions present in a basic aqueous solution with a pOH of 2.4

What have you been asked to do?
Calculate the concentration of hydroxide ions
[OH^-] = ? mol L^-1

What information (data) have you been given?
Extract the data from the question:
pOH = 2.4

What is the relationship between what you know and what you need to find out?
Write the equation (formula) for calculating [OH^-_(aq)]:
[OH^-_(aq)] = 10^-pOH

Substitute the value for pOH into the equation and solve:
[OH^-_(aq)] = 10^-2.4
= 4.0 × 10^-3 mol L^-1

Is your answer plausible?
Check your answer: use your calculated value of [OH^-_(aq)] to calculate the pOH and make sure it is 2.4:
pOH = -log₁₀[OH^-] = -log₁₀[4.0 × 10^-3] = 2.4

State your solution to the problem:
[OH^-_(aq)] = 4.0 × 10^-3 mol L^-1

Question 2. Calculate the concentration of barium ions present in a solution of barium hydroxide with a pOH of 1.7

What have you been asked to do?
Calculate the concentration of barium ions
[Ba²⁺] = ? mol L^-1

What information (data) have you been given?
Extract the data from the question:
pOH = 1.7

What is the relationship between what you know and what you need to find out?
Write the equation (formula) for calculation [OH^-_(aq)]:
[OH^-_(aq)] = 10^-pOH
Substitute in the value for pOH and solve:
[OH^-_(aq)] = 10^-1.7
= 0.020 mol L^-1

Write the balanced chemical equation for the dissociation of barium hydroxide in water:
Ba(OH)₂ → Ba²⁺_(aq) + 2OH^-_(aq)

Determine the stoichiometric ratio (mole ratio) of barium ions to hydroxide ions in solution:
Ba²⁺_(aq) : OH^-_(aq)
1 : 2 or
½ : 1

Use the stoichiometric ratio (mole ratio) to determine the concentration of barium ions in solution:
moles Ba²⁺_(aq) = ½ × moles OH^-_(aq)
moles(Ba²⁺_(aq))/volume of solution = ½ × moles(OH^-_(aq))/volume of solution
[Ba²⁺_(aq)] = ½ × [OH^-_(aq)]

Substitute in the values and solve:
[Ba²⁺_(aq)] = ½ × [OH^-_(aq)]
= ½ × 0.020
= 0.010 mol L^-1

Is your answer plausible?
Check your answer: use your calculated value of [Ba²⁺] to find the pOH of the solution and make sure it is 1.7
[Ba²⁺] = 0.010 mol L^-1 so the concentration of [OH^-] = 2 × 0.010 = 0.020 mol L^-1
pOH = -log₁₀[OH^-] = -log₁₀[0.020] = 1.7

State your solution to the problem:
[Ba²⁺_(aq)] = 0.010 mol L^-1

Question 3. Calculate the mass of potassium hydroxide that needs to be dissolved in water to make 250 mL of aqueous solution with a pOH of 2.3

What have you been asked to do?
Calculate the mass of potassium hydoxide
mass(KOH) = ? g

What information (data) have you been given?
Extract the data from the question:
pOH = 2.3
volume of solution, V = 250 mL
V = 250 mL ÷ 1000 mL/L
= 0.250 L

What is the relationship between what you know and what you need to find out?
Use the pOH to calculate the concentration of hydroxide ions in solution:
Write the equation for calculating [OH^-_(aq)]:
[OH^-_(aq)] = 10^-pOH
Substitute in the value of the pOH and solve:
[OH^-_(aq)] = 10^-2.3 = 5.01 × 10^-3 mol L^-1
Use the volume of the solution to calculate the moles of hydroxide ions present in the solution:
concentration of hydroxide ions, c(OH^-_(aq)) = 5.01 × 10^-3 mol L^-1
volume of solution, V = 0.250 L
moles of hydroxide ions, n(OH^-_(aq)) = ? mol
Since c = n ÷ V (concentration in mol L^-1 = moles ÷ volume in litres)
n = c × V (moles = concentration in mol L^-1 × volume in litres)
So
n(OH^-_(aq)) = c(OH^-_(aq)) × V
= 5.01 × 10^-3 × 0.250
= 1.25 × 10^-3 mol

Calculate the moles of potassium hydroxide, n(KOH):
KOH is a strong base so it dissociates completely in water.
Write the dissociation equation:
KOH → K⁺_(aq) + OH^-_(aq)
Use the dissociation equation to find the stoichiometric ratio (mole ratio),
n(KOH) : n(OH^-_(aq)) is 1 : 1
So, n(KOH) = n(OH^-_(aq)) = 1.25 × 10^-3 mol

Calculate the mass of potassium hydroxide, KOH:
moles = mass in grams ÷ molar mass in g mol^-1
So, mass in grams = moles × molar mass in g mol^-1
moles of KOH = 1.25 × 10^-3 mol
molar mass of KOH = 39.10 + 16.00 + 1.008 = 56.108 g mol^-1
(using the Periodic Table to find the molar mass each element present in KOH)

mass of KOH = 1.25 × 10^-3 ~~mol~~ × 56.108 g ~~mol^-1~~
= 0.070 g

Is your answer plausible?
Check your answer by using your calculated mass of KOH to find the concentration of OH^- and then check that the pOH of the solution is indeed 2.3
moles KOH = mass/molar mass = 0.070/56.108 = 1.248 × 10^-3 moles
moles OH^- = moles KOH = 1.248 × 10^-3 moles
[OH^-] = moles/volume = 1.248 × 10^-3 moles/0.250 L = 4.992 × 10^-3 mol L^-1
pOH = -log₁₀[OH^-] = -log₁₀[4.992 × 10^-3] = 2.3

State your solution to the problem:
mass of KOH = 0.070 g

Can you apply this?

Join AUS-e-TUTE!

Take the exam now!

1. The concentration of an undissociated strong base in aqueous in solution is a theoretical quantity since in practice there are no 'molecules' of undissociated base in the solution.
Chemists find it useful to talk about the concentration of sodium hydroxide solutions for instance, and will write this as [NaOH_(aq)], but it is understood that the solution contains Na⁺_(aq) and OH^-_(aq) but NOT NaOH_(aq).