Calculating Equilibrium Concentrations
- Page ID
- 1316
\(K_a\) is an acid dissociation constant, also known as the acid ionization constant. It describes the likelihood of the compounds and the ions to break apart from each other. As we already know, strong acids completely dissociate, whereas weak acids only partially dissociate. A big \(K_a\) value will indicate that you are dealing with a very strong acid and that it will completely dissociate into ions. A small \(K_a\) will indicate that you are working with a weak acid and that it will only partially dissociate into ions.
General Guide to Solving Problems involving \(K_a\)
Generally, the problem usually gives an initial acid concentration and a \(K_a\) value. From there you are expected to know:
- How to write the \(K_a\) formula
- Set up in an ICE table based on the given information
- Solve for the concentration value, x.
- Use x to find the equilibrium concentration.
- Use the concentration to find pH.
How to write the \(K_a\) formula
The general formula of an acid dissociating into ions is
\[HA_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + A^-_{(aq)} \label{1}\]
with
- \(HA\) is the acid,
- \(A^-\) is the conjugate base and
- \(H_3O^+\) is the the hydronium ion
By definition, the \(K_a\) formula is written as the products of the reaction divided by the reactants of the reaction
\[K_a = \dfrac{[Products]}{[Reactants]} \label{2}\]
Based off of this general template, we plug in our concentrations from the chemical equation. The concentrations on the right side of the arrow are the products and the concentrations on the left side are the reactants. Using this information, we now can plug the concentrations in to form the \(K_a\) equation. We then write:
\[K_a = \dfrac{[H_3O^+][A^-]}{[HA]} \label{3}\]
The concentration of the hydrogen ion (\([H^+]\)) is often used synonymously with the hydrated hydronium ion (\([H_3O^+]\)).
To find a concentration of hydronium ions in solution from a pH, we use the formula:
\[[H_3O^+]= 10^{-pH}\]
This can be flipped to calculate pH from hydronium concentration:
\[pH = -\log[H_3O^+]\]
- An acidic solution is one that has an excess of \(H_3O^+\) ions compared to \(OH^-\) ions.
- An basic (or alkaline) solution is one that has an excess of \(OH^-\) ions compared to \(H_3O^+\) ions.
- A neutral solution is one that has equal concentrations of \(OH^-\) ions and \(H_3O^+\) ions.
At 25 °C, we can correlate whether a solution is acidic, basic, or neutral based off of the measured pH of the solutions:
- pH = 7 is neutral
- pH > 7 is basic
- pH < 7 is acidic
However, these relationships are not valid at temperatures outside 25 °C.
Example \(\PageIndex{1}\): Concentrated Solution of Hypobromous Acid
Calculate the pH of a weak acid solution of 0.2 M HOBr, given:
\[HOBr + H_2O \rightleftharpoons H_3O^+ + OBr^-\]
\[K_a = 2 \times 10^{-9}\]
SOLUTION
Step 1: The ICE Table
Since we were given the initial concentration of HOBr in the equation, we can plug in that value into the Initial Concentration box of the ICE chart. Considering that no initial concentration values were given for H_{3}O^{+} and OBr-, we can assume that none was present initially, and we indicate this by placing a zero in the corresponding boxes. M stands for molarity.
HOBr | H_{3}O^{+} | OBr^{-} | |
---|---|---|---|
Initial Concentration | 0.2 M | 0 | 0 |
Change in Concentration | |||
Equilibrium Concentration |
Because we started off without an initial concentration of H_{3}O^{+} and OBr^{-}, it has to come from somewhere. In the Change in Concentration box, we add a +x because while we do not know what the numerical value of the concentration is at the moment, we do know that it has to be added and not taken away. In contrast, since we did start off with a numerical value of the initial concentration, we know that it has to be taken away to reach equilibrium. Because of this, we add a -x in the HOBr box.
HOBr | H_{3}O^{+} | OBr^{-} | |
---|---|---|---|
Initial Concentration | 0.2 M | 0 | 0 |
Change in Concentration | -x M | +x M | +x M |
Equilibrium Concentration |
Now its time to add it all together! Go from top to bottom and add the Initial concentration boxes to the Change in concentration boxes to get the Equilibrium concentration.
HOBr | H_{3}O^{+} | OBr^{-} | |
---|---|---|---|
Initial Concentration | 0.2 M | 0 | 0 |
Change in Concentration | -x M | +x M | +x M |
Equilibrium Concentration | (0.2 - x) M | x M | x M |
Step 2: Create the \(K_a\) equation using this equation: \(K_a = \dfrac{[Products]}{[Reactants]}\)
\(K_a = \dfrac{[H_3O^+][OBr-]}{[HOBr-]}\)
Step 3: Plug in the information we found in the ICE table
\(K_a = \dfrac{(x)(x)}{(0.2 - x)}\)
Step 4: Set the new equation equal to the given Ka
\[2 \times 10^{-9} = \dfrac{(x)(x)}{(0.2 - x)}\]
Step 5: Solve for x
\[x^2 + (2 \times 10^{-9})x - (4 \times 10^{-10}) = 0\]
To solve for x, we use the quadratic formula
- \(a=1\)
- \(b=2 \times 10^{-9}\)
- \(c=-4 \times 10^{-10}\)
\[x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}= \dfrac{-2 \times 10^{-9} \pm \sqrt{(2 \times10^{-9})^2 - 4(1)(-4 \times 10^{-10})}}{2(1)}\]
Step 6: Plug x back into the ICE table to find the concentration
\[x= [H_3O^+] = 2 \times 10^{-5} \; M\]
Step 7: Use the formula using the concentration to find pH
\[pH = -\log[H_3O^+] = -\log(2 \times 10^{-5}) = -(-4.69) = 4.69\]
\[pH= 4.69\]
Example \(\PageIndex{2}\): Concentrated Solution of Acetic Acid (Vineger)
For acetic acid, HC_{2}H_{3}O_{2}, the \(K_a\) value is \(1.8 \times 10^{-5}\). Calculate the concentration of H_{3}O^{+} in a 0.3 M solution of HC_{2}H_{3}O_{2}_{.}
SOLUTION
Step 1: The ICE Table
Since we were given the initial concentration of HC_{2}H_{3}O_{2} in the original equation, we can plug in that value into the Initial Concentration box of the ICE chart. Considering that no initial concentration values were given for \(H_3O^+\) and \(C_2H_3O_2^-\), we assume that none was present initially, and we indicate this by placing a zero in the corresponding boxes.
HC_{2}H_{3}O_{2} | H_{3}O^{+} | C_{2}H_{3}O_{2} | |
---|---|---|---|
Initial Concentration | 0.3 M | 0 | 0 |
Change in Concentration | |||
Equilibrium Concentration |
Because we started off without any initial concentration of H_{3}O^{+} and C_{2}H_{3}O_{2}-, is has to come from somewhere. For the Change in Concentration box, we add a +x because while we do not know what the numerical value of the concentration is at the moment, we do know that it has to be added and not taken away. In contrast, since we did start off with a numerical value of the initial concentration, we know that it has to be taken away to reach equilibrium. Because of this, we add a -x in the \(HC_2H_3O_2\) box.
HC_{2}H_{3}O_{2} | H_{3}O^{+} | C_{2}H_{3}O_{2} | |
---|---|---|---|
Initial Concentration | 0.3 M | 0 | 0 |
Change in Concentration | -x M | +x M | +x M |
Equilibrium Concentration |
Now its time to add it all together! Go from top to bottom and add the Initial concentration boxes to the Change in concentration boxes to get the Equilibrium concentration.
HC_{2}H_{3}O_{2} | H_{3}O^{+} | C_{2}H_{3}O_{2} | |
---|---|---|---|
Initial Concentration | 0.3 M | 0 | 0 |
Change in Concentration | -x M | +x M | +x M |
Equilibrium Concentration | (0.3 - x) M | x M | x M |
Step 2: Create the \(K_a\) equation using this equation: \(K_a = \dfrac{[Products]}{[Reactants]}\)
\(K_a = \dfrac{[H_3O^+][C_2H_3O_2]}{[HC_2H_3O_2]}\)
Step 3: Plug in the information we found in the ICE table
\[K_a = \dfrac{(x)(x)}{(0.3 - x)}\]
Step 4: Set the new equation equal to the given K_{a}
\[1.8 x 10^{-5} = \dfrac{(x)(x)}{(0.3 - x)}\]
Step 5: Solve for x
\[(x^2)+ (1.8 \times 10^{-5}x)-(5.4 \times 10^{-6})\]
To solve for x, we use the quadratic formula
- \(a=1\)
- \(b=1.8 \times 10^{-5}\)
- \(c=-5.4 \times 10^{-6}\)
\[x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}= \dfrac{-1.8 \times 10^{-5} \pm \sqrt{(1.8 \times10^{-5})^2 - 4(1)(-5.4 \times 10^{-6})}}{2(1)}\]
Step 6: Plug x back into the ICE table to find the concentration
\[x= [H_3O^+] = 0.0023\; M\]
Example \(\PageIndex{3}\): Concentrated Solution of Benzoic Acid
Find the equilibrium concentration of HC_{7}H_{5}O_{2}from a 0.43 M solution of Benzoic Acid, HC_{7}H_{5}O_{2}_{. }
SOLUTION
Given: \(K_a\) for HC_{7}H_{5}O_{2}= 6.4 x 10^{-5 }
Step 1: The ICE Table
HC_{7}H_{5}O_{2} | H_{3}O^{+} | C_{7}H_{5}O_{2}^{-} | |
---|---|---|---|
Initial Concentration | 0.43 M | 0 | 0 |
Change in Concentration | -x | +x | +x |
Equilibrium Concentration | (0.43-x)M | x M | x M |
Step 2: Create the \(K_a\) equation using this equation :\(K_a = \dfrac{[Products]}{[Reactants]}\)
\(K_a = \dfrac{[H_3O^+][C_7H_5O_2-]}{[HC_7H_5O_2]}\)
Step 3: Plug in the information we found in the ICE table
\(K_a = \dfrac{(x)(x)}{(0.43 - x)}\)
Step 4: Set the new equation equal to the given Ka
\(6.4 x 10^{-5} = \dfrac{(x)(x)}{(0.43 - x)}\)
Step 5: Solve for x.
Step 6: Plug x back into the ICE table to find the concentration
[HC_{7}H_{5}O_{2}]= (0.43-x)M
[HC_{7}H_{5}O_{2}]= (0.43-0.0052)M
Answer: [HC_{7}H_{5}O_{2}]= 0.425 M
Example \(\PageIndex{4}\): Concentrated Solution of Hypochlorous acid
For a 0.2 M solution of Hypochlorous acid, calculate all equilibrium concentrations.
SOLUTION
Given: \(K_a = 3.5 \times 10^{-8}\)
Step 1: The ICE Table
HOCl | H_{3}O^{+} | OCl^{-} | |
---|---|---|---|
Initial Concentration | 0.2 M | 0 | 0 |
Change in Concentration | -x M | +x M | +x M |
Equilibrium Concentration | (0.2 - x) M | x M | x M |
Step 2: Create the \(K_a\) equation using this equation: \(K_a = \dfrac{[Products]}{[Reactants]}\)
\(K_a = \dfrac{[H_3O^+][OCl-]}{[HOCl-]}\)
Step 3: Plug in the information we found in the ICE table
\(K_a = \dfrac{(x)(x)}{(0.2 - x)}\)
Step 4: Set the new equation equal to the given Ka
\(3.5 x 10^{-8} = \dfrac{(x)(x)}{(0.2 - x)}\)
Step 5: Solve for x
Step 6: Plug x back into the ICE table to find the concentration
[HOCl]= [(.2)-(8.4 x 10^{-5})]=.199
[H_{3}O+]=8.4 x 10^{-5}
[OCl-]=8.4 x 10^{-5}
Example \(\PageIndex{5}\): pH
Calculate the pH from the equilibrium concentrations of [H_{3}O^{+}] in Example \(\PageIndex{4}\).
SOLUTION
Given:
[HOCl]=0.199
[H_{3}O^{+}]=8.4 x 10^{-5}
[OCl-]=8.4 x 10^{-5}
Step 1: Use the formula using the concentration of [H_{3}O^{+}] to find pH
\[pH = -\log[H3O+] = -\log(8.4 x 10^{-5}) = 4.08\]
References
- Petrucci, et al. General Chemistry: Principles & Modern Applications; Ninth Edition. Pearson/Prentice Hall; Upper Saddle River, New Jersey 07.
- Hawkes, Stephen J. "Easy Derivation of pH (pK_{a1} + pK_{a2}) / 2 Using Autoprotolysis of HA^{-}: Doubtful Value of the Supposedly More Rigorous Equation." J. Chem. Educ. 2000 77 1183.
- van Lubeck, Henk. "Why Not Replace pH and pOH by Just One Real Acidity Grade, AG?." J. Chem. Educ. 1999 76 892.
Contributors
- Kellie Berman (UCD), Alysia Kreitem (UCD)