# Hydroxide Ion Concentration Calculations Chemistry Tutorial

## Key Concepts

• The hydroxide ion concentration in an aqueous solution, [OH-], in mol L-1, can be calculated if the pOH of the solution is known.
• pOH is defined as the negative logarithm (to base 10) of the hydroxide ion concentration in mol L-1
pOH = -log10[OH-]
[OH-] is the concentration of hydroxide ions in mol L-1 (mol/L or M)
• [OH-] in mol L-1 can be calculated using the equation (formula):
[OH-] = 10-pOH
• As the pOH increases, the concentration of hydroxide ions in the solution decreases.

As the pOH decreases, the concentration of hydroxide ions in the solution increases.

No ads = no money for us = no free stuff for you!

## [OH-] Calculations Using a Calculator

You will need to locate the 10x button on your calculator.
This button will probably be labelled 10x and be above the button labelled log or LOG
For calculations involving pOH, do NOT use the button labelled ex (above the button labelled ln or LN)

On AUS-e-TUTE's calculator the 10x button is above the log button positioned in the top left hand corner of the calculator.

To use the 10x button to calculate [OH-] you will need to:

1. Enter the number (pOH) in the
2. Multiply the pOH by -1 to find -pOH
3. Click the INV button
4. Click the log button

On the calculator shown below, we have disabled all the buttons except the INV and log button.

 To find the [OH-] in mol L-1 of an aqueous solution given the pOH of the solution: First, we need to multiply the pOH value by -1. Enter a pOH value between 0 and 14 in the textbox: pOH = Click the -pOH button: = -1 × pOH = -1 × = Now click the INV button on the calculator shown on the right [OH-] = 10-pOH = 10 Then click the log button on the calculator shown on the right [OH-] = Click the Reset button to clear these numbers and start a new calculation:

Check the calculations below:

 pOH [OH-] mol L-1 trend 3 2.3 2 1.3 1 0.3 0 -0.7 -1 1 x 10-3 5 x 10-3 1 x 10-2 5 x 10-2 1 x 10-1 5 x 10-1 1 x 100 5 x 100 1 x 101 (= 0.001) (= 0.005) (= 0.01) (= 0.05) (= 0.1) (0.5) (= 1) (= 5) (= 10) decreasing pOH → increasing hydroxide ion concentration →

Note the following relationships between the concentration of hydroxide ions and the pH:

• Decreasing pOH increases [OH-]
• Increasing pOH decreases [OH-]

Notice that a change of 1 pOH unit results in a tenfold change in the concentration of hydroxide ions results:

 pOH = 3 - 1 = pOH = 2 - 1 = pOH = 1 - 1 = pOH = 0 - 1 = pOH = -1 [OH-]=1 x 10-3 M x10= [OH-]=1 x 10-2 M x10= [OH-]=1 x 10-1 M x10= [OH-]=1 x 100 M x10= [OH-]=1 x 101 M

 pOH = 2.3 - 1 = pOH = 1.3 - 1 = pOH = 0.3 - 1 = pOH = -0.7 [OH-]=5 x 10-3 M x10= [OH-]=5 x 10-2 M x10= [OH-]=5 x 10-1 M x10= [OH-]=5 x 100 M

Do you know this?

Play the game now!

## [OH-] Graphs

Using the formula (equation) [OH-] = 10-pOH we can calculate the [OH-] in mol L-1 of solutions with varying pOH values and plot these values on a graph.

An example is shown below:

Data Graph Trends
pOH 10-pOH = [OH-]
mol L-1
0.00 100.00 = 1.00
0.05 10-0.05 = 0.89
0.10 10-0.10 = 0.79
0.15 10-0.15 = 0.71
0.20 10-0.20 = 0.63
0.25 10-0.25 = 0.56
0.30 10-0.30 = 0.50
0.35 10-0.35 = 0.45
0.40 10-0.40 = 0.40
0.45 10-0.45 = 0.35
 [OH-] mol L-1 Relationship Between [OH-] and pOHpOH
Increasing pOH decreases [OH-].

Higher pOH gives a lower [OH-].

Decreasing pOH increases [OH-].

Lower pOH gives a higher [OH-].

If you place your mouse over any of the points in the graph, a small box should appear to show you the values of the hydroxide ion concentration and the pOH at that point.

Do you understand this?

Take the test now!

## Worked Examples

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

Question 1. The pOH of an aqueous solution of a base is determined to be 4.4
What is the concentration of hydroxide ions in mol L-1 in this basic solution?

1. What have you been asked to do?
Calculate concentration of hydroxide ions in mol L-1
[OH-(aq)] = ? mol L-1
2. What information (data) have you been given?
Extract the data from the question:
pOH = 4.4
3. What is the relationship between what you know and what you need to find out?
Write the equation (formula) for finding [OH-] given the pOH:
[OH-] = 10-pOH
4. Substitute in the values and solve:
[OH-] = 10-pOH
[OH-] = 10-4.4 = 3.98 × 10-5 mol L-1
Use your calculated value of [OH-] to find pOH and compare your value to that given in the question:
pOH = -log10[OH-] = -log10[3.98 × 10-5] = 4.4
Since this value is the same as that given in the question, we are confident our answer is correct.
6. State your solution to the problem:
[OH-(aq)] = 3.98 × 10-5 mol L-1

Question 2. The pOH of an aqueous solution of sodium hydroxide is 3.5
Calculate the moles of hydroxide ions in 0.25 L of this solution.

1. What have you been asked to do?
Calculate moles of hydroxide ions
n(OH-(aq)) = ? mol
2. What information (data) have you been given?
Extract the data from the question:
pH = 3.5
volume = 0.25 L
3. What is the relationship between what you know and what you need to find out?
Find the concentration of hydroxide ions:
[OH-] = 10-pOH
= 10-3.5
= 3.16 × 10-4 mol L-1

Write the equation (formula) that relates moles, volume and concentration (molarity):
molarity = moles ÷ volume (L)

Rearrange this equation (formula) to find moles:
moles = molarity (mol L-1) × volume (L)

4. Calculate moles of OH-:
molarity = 3.16 × 10-4 mol L-1
volume = 0.25 L

moles OH- = 3.16 × 10-4 × 0.25
= 7.9 × 10-5 mol

Use your calculated value of moles of OH- to find pOH and compare your value to that given in the question:
[OH-] = moles ÷ volume = (7.9 × 10-5 ) ÷ 0.25 = 3.16 × 10-4 mol L-1
pOH = -log10[OH-] = -log10[3.16 × 10-4] = 3.5
Since this value is the same as that given in the question, we are confident our answer is correct.
6. State your solution to the problem:
n(OH-(aq)) = 7.9 × 10-5 mol

Question 3. An aqueous solution of ammonium hydroxide has a pOH of 6.0
Calculate the number of hydroxide ions present in 100 mL of this solution.

1. What have you been asked to do?
Calculate number of hydroxide ions
N(OH-(aq)) = ?
2. What information (data) have you been given?
Extract the data from the question:
pOH = 6.0
volume = 100 mL
3. What is the relationship between what you know and what you need to find out?
Find the concentration of hydroxide ions:
[OH-] = 10-pOH = 10-6.0 mol L-1

Find the moles of OH-:
Write the equation (formula) that relates moles, volume and concentration (molarity):
molarity = moles ÷ volume (L)

Rearrange this equation (formula) to find moles:
moles = molarity (mol L-1) × volume (L)

Calculate moles of OH-:
molarity = 10-6 mol L-1
volume = 100 mL
Convert volume in mL to L by dividing by 1000:
volume = 100 mL ÷ 1000 mLL-1 = 0.10 L

moles OH- = 10-6 × 0.10 = 10-7 mol

4. Calculate the number of hydroxide ions:
1 mole = Avogadro's Number of particles = 6.02 x 1023 particles

10-7 mol OH- = 10-7 × ( 6.02 × 1023)
= 6.0 × 1016 hydroxide ions

Use your calculated value of number of OH- particles to find pOH and compare your value to that given in the question:
moles = N ÷ NA = (6.0 × 1016) ÷ ( 6.02 × 1023) = 10-7 mol
[OH-] = moles ÷ volume in litres = 10-7 ÷ 100mL/1000mL/L = 10-6 mol L-1
pOH = -log10[OH-] = -log10[10-6] = 6
Since this value is the same as that given in the question, we are confident our answer is correct.
6. State your solution to the problem:
N(OH-(aq)) = 6.0 × 1016

Question 4. An aqueous solution of potassium hydroxide contains 0.0005 moles of hydroxide ions and has a pOH of 4.3.
Calculate the volume of the solution in litres.

1. What have you been asked to do?
Calculate volume of the solution in litres
V = ? L
2. What information (data) have you been given?
Extract the data from the question:
pOH = 4.3
moles OH- = 0.0005 mol
3. What is the relationship between what you know and what you need to find out?
Find the concentration of hydroxide ions:
[OH-] = 10-pOH
= 10-4.3
= 5.0 × 10-5 mol L-1

Write the equation (formula) that relates moles, volume and concentration (molarity):
molarity = moles ÷ volume (L)

Rearrange this equation (formula) to find volume:
volume= moles ÷ molarity (mol L-1)

4. Calculate volume of solution:
moles OH- = 0.0005 mol
molarity = 5.0 × 10-5 mol L-1

volume of solution = 0.0005 ÷ (5.0 × 10-5)
= 10 L

Use your calculated value of volume of solution to find pOH and compare your value to that given in the question:
[OH-] = moles ÷ volume in litres = 0.0005 ÷ 10 = 5 × 10-5 mol L-1
pOH = -log10[OH-] = -log10[5 × 10-5] = 4.3
Since this value is the same as that given in the question, we are confident our answer is correct.
6. State your solution to the problem:
V = 10 L

Can you apply this?

Join AUS-e-TUTE!

Do the drill now!