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Mobile App Chapter 16: Problem 26
A \(0.23 \mathrm{M}\) solution of a weak acid (HA) has a pH of \(2.89\). What is the \(K_{\mathrm{a}}\) for this acid?
Short Answer
Expert verified
The \(K_{\text{a}}\) value for the acid is \(7.28 \times 10^{-6}\).
Step by step solution
01
Understand the pH Definition
The pH of a solution is a measure of its hydrogen ion concentration. Given the pH, calculate the hydrogen ion concentration \(\text{[H}^+\text{]}\). The formula to use is \[ \text{[H}^+\text{]} = 10^{-\text{pH}} \] Substituting the given pH value: \(\text{[H}^+\text{]} = 10^{-2.89} = 1.29 \times 10^{-3} \) M.
02
Write the Ionization Equation
For the weak acid HA dissociating in water, the reaction is: \(\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-\). Since \[ \text{[H}^+ \text{]} = 1.29 \times 10^{-3} \] M, we also assume \[ \text{[A}^- \text{]} = 1.29 \times 10^{-3} \] M because each HA molecule produces one H\textsuperscript{+} and one A\textsuperscript{-} ion.
03
Calculate the Remaining HA Concentration
Determine the initial concentration of HA and the change in HA concentration due to dissociation: \[ \text{Initial [HA]} = 0.23 \text{ M} \] Given that \[ \text{[HA dissociating]} = 1.29 \times 10^{-3} \] M, the remaining concentration of HA is: \[ \text{[HA]}_{\text{final}} = 0.23 - 1.29 \times 10^{-3} = 0.2287 \] M.
04
Use the Equilibrium Expression
The equilibrium constant for the dissociation of HA, \(K_{\text{a}}\), is given by: \[ K_{\text{a}} = \frac{\text{[H}^+] \times \text{[A}^-]}{\text{[HA]}} \] Substitute the known values: \[ K_{\text{a}} = \frac{(1.29 \times 10^{-3}) \times (1.29 \times 10^{-3})}{0.2287} \]
05
Calculate the Ka Value
Calculate the numerical value: \[ K_{\text{a}} = \frac{(1.29 \times 10^{-3})^2}{0.2287} = \frac{1.6641 \times 10^{-6}}{0.2287} = 7.28 \times 10^{-6} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
pH and Hydrogen Ion Concentration
The pH of a solution tells us how acidic or basic it is. It directly relates to the concentration of hydrogen ions (H\textsuperscript{+}) in the solution. The lower the pH, the higher the concentration of hydrogen ions. To find the concentration of H\textsuperscript{+} ions from pH, you can use the formula: \[ \text{[H}^+\text{]} = 10^{-\text{pH}} \]
For instance, if the pH is 2.89,\ the hydrogen ion concentration \text{[H}^+\text{]} is \[ 10^{-2.89} = 1.29 \times 10^{-3} \text{ M} \] This formula helps convert the pH value to an actual concentration number for calculations.
For instance, if the pH is 2.89,\ the hydrogen ion concentration \text{[H}^+\text{]} is \[ 10^{-2.89} = 1.29 \times 10^{-3} \text{ M} \] This formula helps convert the pH value to an actual concentration number for calculations.
Ionization of Weak Acids
Weak acids only partially ionize in water. This means a fraction of the acid molecules dissociate into ions. For a weak acid HA, the ionization in water can be shown as:
\[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \]
Given that the hydrogen ion concentration \text{[H}^+\text{]} is equal to \text{[A}^-\text{]}, we can know that each molecule of HA that dissociates produces one \text{H}^+ and one \text{A}^-. If \[ \text{[H}^+\text{]}=1.29 \times 10^{-3} \text{ M} \] then \text{[A}^-\text{]} is also \[ 1.29 \times 10^{-3} \text{ M} \]
\[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \]
Given that the hydrogen ion concentration \text{[H}^+\text{]} is equal to \text{[A}^-\text{]}, we can know that each molecule of HA that dissociates produces one \text{H}^+ and one \text{A}^-. If \[ \text{[H}^+\text{]}=1.29 \times 10^{-3} \text{ M} \] then \text{[A}^-\text{]} is also \[ 1.29 \times 10^{-3} \text{ M} \]
Acid Dissociation Constant (Ka)
The acid dissociation constant, represented as \text{K_a}, is a measure of the strength of a weak acid. It quantifies the extent to which an acid can ionize in solution. For an acid HA, the expression for \text{K_a} is given by: \[ K_{\text{a}} = \frac{\text{[H}^+] \times \text{[A}^-]}{\text{[HA]}} \]
Using values from our example:
\[ K_{\text{a}} = \frac{(1.29 \times 10^{-3}) \times (1.29 \times 10^{-3})}{0.2287} \]
this results in: \[ K_{\text{a}} = 7.28 \times 10^{-6} \] A higher \text{K_a} means a stronger weak acid, because it implies greater ionization.
Using values from our example:
\[ K_{\text{a}} = \frac{(1.29 \times 10^{-3}) \times (1.29 \times 10^{-3})}{0.2287} \]
this results in: \[ K_{\text{a}} = 7.28 \times 10^{-6} \] A higher \text{K_a} means a stronger weak acid, because it implies greater ionization.
Chemical Equilibrium
Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction. For weak acids in water, equilibrium involves the dissociation of the acid and the recombination of ions. In the equation
\[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \]
The concentration of reactants (HA) and products (H\textsuperscript{+} and A\textsuperscript{-}) remain constant over time. At equilibrium, we can apply the equilibrium expression, substituting the equilibrium concentrations to find \text{K_a}.
\[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \]
The concentration of reactants (HA) and products (H\textsuperscript{+} and A\textsuperscript{-}) remain constant over time. At equilibrium, we can apply the equilibrium expression, substituting the equilibrium concentrations to find \text{K_a}.
Solution Concentration Calculations
Calculating solution concentrations involves determining the amount of solute in a given volume of solution. For our weak acid solution, the initial concentration of HA was 0.23 M. Given the ionization, the remaining concentration of HA after dissociation can be calculated as: \[ \text{[HA]}_{\text{remaining}} = 0.23 \text{ M} - 1.29 \times 10^{-3} \text{ M} = 0.2287 \text{ M} \] It's essential to know the initial concentration and how much the acid ionizes to find the equilibrium concentrations necessary for \text{K_a} calculations.
