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Calculating the pOH of Strong Monoprotic Acids
Before the acid is added to water, water molecules are in equilibrium with hydrogen ions and hydroxide ions:
water | | hydrogen ions | + | hydroxide ions |
H_{2}O | | H^{+}_{(aq)} | + | OH^{-}_{(aq)} |
Only a very small number of water molecules dissociate into H^{+}_{(aq)} and OH^{-}_{(aq)}.
At 25°C, [H^{+}_{(aq)}] = [OH^{-}_{(aq)}] ≈ 10^{-7} mol L^{-1} (a very low concentration)
and K_{w} = [H^{+}_{(aq)}][OH^{-}_{(aq)}] = 10^{-7} × 10^{-7} = 10^{-14}
By calculating the logarithm (to base 10) of all the species above, we change the calculation to an addition instead of a multiplication:
K_{w} = [H^{+}_{(aq)}][OH^{-}_{(aq)}]
log_{10}(K_{w}) = log_{10}[H^{+}_{(aq)}] + log_{10}[OH^{-}_{(aq)}]
If we then multiply throughout by -1:
-log_{10}(K_{w}) = -log_{10}[H^{+}_{(aq)}] + -log_{10}[OH^{-}_{(aq)}]
We can see that:
-log_{10}(K_{w}) = pH + pOH
since pH = -log_{10}[H^{+}_{(aq)}]
and pOH = -log_{10}[OH^{-}_{(aq)}]
For water at 25^{o}C,
K_{w} = 10^{-14}
log_{10}K_{w} = log_{10}10^{-14} = -14
-log_{10}K_{w} = -log_{10}10^{-14} = 14
so, 14 = pH + pOH
When a strong monoprotic acid is added to water, the acid dissociates completely to form H^{+}_{(aq)} and an aqueous anion:
acid | → | hydrogen ions | + | anions |
HA | → | H^{+}_{(aq)} | + | A^{-}_{(aq)} |
Adding the acid to the water disturbs the water dissociation equilibrium:
H_{2}O H^{+}_{(aq)} + OH^{-}_{(aq)}
By Le Chatelier's Principle, adding more H^{+}_{(aq)} to the water will shift the equilibrium position to the left.
The water dissociation equilibrium system responds to the addition of more H^{+}_{(aq)} by reacting some of the H^{+}_{(aq)} with some of the OH^{-}_{(aq)} in order to re-establish equilibrium.
So, increasing the concentration of H^{+}_{(aq)} in the water, reduces the concentration of OH^{-}_{(aq)}, but, the water dissociation constant does not change^{4}, K_{w} is still 10^{-14}.
So, K_{w} = [H^{+}_{(aq)}][OH^{-}_{(aq)}] = 10^{-14}
and -log_{10}K_{w} = 14 = pH + pOH
[H^{+}_{(aq)}] mol L^{-1} |
[OH^{-}_{(aq)}] mol L^{-1} |
K_{w} |
At 25^{o}C |
pH |
pOH |
-log(K_{w}) (pK_{w}) |
10^{-7} |
10^{-7} |
10^{-14} |
pure water (before acid added) |
7 |
7 |
14 |
> 10^{-7} |
< 10^{-7} |
10^{-14} |
aqueous solution (after acid added) |
< 7 |
> 7 |
14 |
We can use the value of K_{w} or -log_{10}K_{w} (pK_{w}) and pH or [H^{+}_{(aq)}] to calculate pOH at a given temperature:
If we know the pH of the acidic solution, we can calculate the pOH:
pH + pOH = -log_{10}K_{w}
So, pOH = -log_{10}K_{w} - pH
At 25°C : pOH = 14 - pH
If we know the concentration of hydrogen ions in solution, we can calculate the pOH:
K_{w} = [H^{+}_{(aq)}][OH^{-}_{(aq)}]
By rearranging this equation (formula) we can determine the concentration of hydroxide ions in the aqueous solution:
[OH^{-}_{(aq)}] = K_{w} ÷ [H^{+}_{(aq)}]
so, pOH = -log_{10}[OH^{-}_{(aq)}] = -log_{10}(K_{w} ÷ [H^{+}_{(aq)}])
Both [H^{+}_{(aq)}] and [OH^{-}_{(aq)}] must be in units of mol L^{-1} (mol/L or M)
Worked Examples
(based on the StoPGoPS approach to problem solving in chemistry.)
Question 1. Calculate the pOH of an aqueous solution of hydrochloric acid with a hydroxide ion concentration of 1.0 × 10^{-12} mol L^{-1}.
- What have you been asked to do?
Calculate the pOH
pOH = ?
- What information (data) have you been given?
Extract the data from the question:
[OH^{-}_{(aq)}] = 1.0 × 10^{-12} mol L^{-1}
- What is the relationship between what you know and what you need to find out?
Write the equation (formula) for calculating pOH:
pOH = -log_{10}[OH^{-}_{(aq)}]
- Subsitute in the value for the hydroxide ion concentration and solve:
pOH = -log_{10}[1.0 × 10^{-12}]
= 12
- Is your answer plausible?
Use your calculated value for pOH to find [OH^{-}] and compare it to that given in the question:
[OH^{-}] = 10^{-pOH} = 10^{-12}
Since this value is the same as that given in the question we are confident our answer is correct.
- State your solution to the problem:
pOH = 12
Question 2. Calculate the pOH of an aqueous solution of hydrochloric acid with a pH of 2.5 at 25°C.
- What have you been asked to do?
Calculate the pOH
pOH = ?
- What information (data) have you been given?
Extract the data from the question:
pH = 2.5
temperature = 25°C
so, K_{w} = 10^{-14} (from Data Sheet)
and, -log_{10}K_{w} = -log_{10}10^{-14} = 14
- What is the relationship between what you know and what you need to find out?
Write the equation (formula) for calculating pOH:
pH + pOH = 14
Rearrange this equation (formula) to find pOH:
pOH = 14 - pH
- Substitute in the pH value and solve:
pOH = 14 - 2.5
= 11.5
- Is your answer plausible?
Use your calculated value for pOH to find pH and compare it to that given in the question:
pH + pOH = 14
pH + 11.5 = 14
pH = 14 - 11.5 = 2.5
Since this value is the same as that given in the question we are confident our answer is correct.
- State your solution to the problem:
pOH = 11.5
Question 3. Calculate the pOH of 0.025 mol L^{-1} HCl_{(aq)} at 25°C.
(i) Hydrogen ion concentration method:
- What have you been asked to do?
Calculate the pOH
pOH = ?
- What information (data) have you been given?
Extract the data from the question:
[HCl_{(aq)}] = 0.025 mol L^{-1}
temperature = 25^{o}C
so, K_{w} = 10^{-14} (from Data Sheet)
- What is the relationship between what you know and what you need to find out?
Calculate the concentration of hydrogen ions present in solution:
HCl_{(aq)} is a strong monoprotic acid, it dissociates fully in water:
HCl → H^{+}_{(aq)} + Cl^{-}_{(aq)}
The stoichiometric ratio (mole ratio) of HCl to H^{+}_{(aq)} is 1 : 1
so, [HCl_{(aq)}] = [H^{+}_{(aq)}] = 0.025 mol L^{-1}
Write the equation (formula) for calculating the concentration of hydroxide ions in solution:
K_{w} = [H^{+}_{(aq)}][OH^{-}_{(aq)}]
so, [OH^{-}_{(aq)}] = K_{w}/[H^{+}_{(aq)}]
Substitute in the values for K_{w} and [H^{+}_{(aq)}] and solve:
[OH^{-}_{(aq)}] = 10^{-14}/[0.025 mol L^{-1}] = 4.0 × 10^{-13}
Write the equation (formula) for calculating pOH:
pOH = -log_{10}[OH^{-}_{(aq)}]
- Substitute in the hydroxide ion concentration and solve:
pOH = -log_{10}[4.0 × 10^{-13}] = 12.4
- Is your answer plausible?
Use your calculated value for pOH to find [H^{+}] and compare it to that given in the question:
pH = 14 - 12.4 = 1.6
[H^{+}] = 10^{-pH} = 10^{-1.6} = 0.025 mol L^{-1} = [HCl_{(aq)}]
Since this value is the same as that given in the question we are confident our answer is correct.
- State your solution to the problem:
pOH = 12.4
(ii) pH method:
- What have you been asked to do?
Calculate the pOH
pOH = ?
- What information (data) have you been given?
Extract the data from the question:
[HCl_{(aq)}] = 0.025 mol L^{-1}
temperature = 25°C
so, K_{w} = 10^{-14} (from Data Sheet)
- What is the relationship between what you know and what you need to find out?
Calculate the concentration of hydrogen ions present in solution:
HCl_{(aq)} is a strong monoprotic acid, it dissociates fully in water:
HCl → H^{+}_{(aq)} + Cl^{-}_{(aq)}
The stoichiometric ratio (mole ratio) of HCl to H^{+}_{(aq)} is 1 : 1
so, [HCl_{(aq)}] = [H^{+}_{(aq)}] = 0.025 mol L^{-1}
Calculate the pH of the solution:
pH = -log_{10}[H^{+}_{(aq)}] = -log_{10}[0.025] = 1.60
Write the equation (formula) for calculating pOH:
-log_{10}K_{w} = pH + pOH
- Substitute in the values for K_{w} and pH and solve:
-log_{10}10^{-14} = 1.60 + pOH
14 = 1.60 + pOH
pOH = 14 - 1.60
= 12.4
- Is your answer plausible?
Use your calculated value for pOH to find [H^{+}] and compare it to that given in the question:
pH = 14 - 12.4 = 1.6
[H^{+}] = 10^{-pH} = 10^{-1.6} = 0.025 mol L^{-1} = [HCl_{(aq)}]
Since this value is the same as that given in the question we are confident our answer is correct.
- State your solution to the problem:
pOH = 12.4
1. The dissociation constant for water, K_{w}, varies with temperature. At 25^{o}C K_{w} ~10^{-14}.
The dissociation constant, or ionisation constant, for water should be provided to you either in the question or on a data sheet.
If no temperature is specified in a question, assume the temperature is 25^{o}C.
2. Hydrogen ions are being represented here as H^{+}, or H^{+}_{(aq)} for hydrogen ions in water (hydrated hydrogen ions).
You can use H_{3}O^{+} (oxidanium or hydronium or oxonium ion) instead to represented hydrogen ions in aqueous solution.
In this case the dissociation of water is 2H_{2}O H_{3}O^{+} + OH^{-}_{(aq)}
and K_{w} = [H_{3}O^{+}][OH^{-}_{(aq)}] = 10^{-14} (at 25^{o}C).
3. K_{w} = 10^{-14} is only an approximation, but it is one that is widely used for convenience.
4. This assumes that the temperature of the water remains constant.
If the temperature changes, the value of K_{w} changes.