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Chapter 8: Problem 104

Consider an initial \(0.040 M\) hypobromous acid (HOBr) solution at a certain temperature. $$\operatorname{HOBr}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OBr}^{-}(a q)$$ At equilibrium after partial dissociation, its pH is found to be \(5.05 .\) What is the acid ionization constant, \(K_{\mathrm{a}},\) for hypobromous acid at this temperature?

Short Answer

Expert verified
The acid ionization constant, \(K_a\), is approximately \(1.98 \times 10^{-9}\).

Step by step solution

01

Understand the Relationship Between pH and Hydrogen Ion Concentration

The pH of a solution is related to its hydrogen ion concentration \([\text{H}^+]\). We use the formula: \[\text{pH} = -\log_{10}([\text{H}^+])\].Given that the pH is 5.05, we can rearrange this formula to find \([\text{H}^+]\):\[[\text{H}^+] = 10^{-5.05}\].This will give us the concentration of hydrogen ions in the solution.
02

Calculate the Hydrogen Ion Concentration

By plugging in the pH value from Step 1, we calculate:\[[\text{H}^+] = 10^{-5.05} \approx 8.91 \times 10^{-6} M\].This is the equilibrium concentration of hydrogen ions in the solution.
03

Determine Equilibrium Concentrations

The reaction \(\text{HOBr}(aq) \rightleftharpoons \text{H}^+(aq) + \text{OBr}^-(aq)\)shows that for every mole of \([\text{H}^+]\) produced, one mole of \([\text{OBr}^-]\) is formed. Therefore, at equilibrium, \([\text{OBr}^-] = 8.91 \times 10^{-6} M\). "The initial concentration of HOBr was \(0.040 M\), and the change in HOBr concentration is equal to the concentration of hydrogen ions, so \([\text{HOBr}] = 0.040 - 8.91 \times 10^{-6} M\)."
04

Write the Expression for the Acid Ionization Constant \(K_a\)

The expression for the acid ionization constant is given by:\[K_a = \frac{[\text{H}^+][\text{OBr}^-]}{[\text{HOBr}]}\]Substitute the equilibrium concentrations from the previous steps into the expression.
05

Calculate \(K_a\) Value

Using the concentrations calculated, plug them into the \(K_a\) expression:\[K_a = \frac{(8.91 \times 10^{-6})(8.91 \times 10^{-6})}{0.040 - 8.91 \times 10^{-6}}\]With the approximation that \(0.040 - 8.91 \times 10^{-6} \approx 0.040\), we find:\[K_a \approx \frac{(8.91 \times 10^{-6})^2}{0.040}\]Calculate this expression to find \(K_a\):\[K_a \approx 1.98 \times 10^{-9}\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Concentrations
When a weak acid like hypobromous acid ( HOBr) is added to water, it doesn't completely dissociate. Instead, some of the molecules dissociate to form hydrogen ions ([H+]) and hypobromite ions ([OBr鈭�]). The solution reaches a state called equilibrium where the rates of dissociation and recombination of the molecules are equal. This means concentrations of the reactants and products remain constant over time.

At equilibrium, the concentration of [H+] will match the concentration of [OBr鈭�] because they are produced in a 1:1 ratio, as indicated by the balanced equation: HOBr(aq) 鈬� H+(aq) + OBr鈭�(aq). The initial concentration of the acid will decrease by the same amount as the hydrogen ion concentration.
  • Example: If the initial concentration of HOBr is 0.040 M and [H+] at equilibrium is 8.91 脳 10-6 M, then [HOBr] has decreased by this amount.
  • Therefore, the equilibrium concentration of HOBr is (0.040 - 8.91 脳 10-6) M.
Understanding these equilibrium processes helps predict how a weak acid behaves in solution.
Hydrogen Ion Concentration
Hydrogen ion concentration ([H+]) is crucial in determining the acidity of a solution. It indicates how many hydrogen ions are present. In the given problem, we are given the pH value of the solution, which allows us to find [H+].

The pH is calculated using the formula: \[\text{pH} = -\log_{10}([\text{H}^+])\] So, when we rearrange this equation to find [H+], it becomes:\[[\text{H}^+] = 10^{-\text{pH}}\] By plugging in the pH of 5.05:\[[\text{H}^+] = 10^{-5.05} \approx 8.91 \times 10^{-6}\] This value represents the molar concentration of hydrogen ions at equilibrium.
pH
The term pH is a logarithmic measure of the hydrogen ion concentration in a solution. It's crucial for understanding the acidity of a solution, as it transforms the difficult-to-read hydrogen ion concentrations into a more manageable scale.

Recognized by the equation:\[\text{pH} = -\log_{10}([\text{H}^+])\]pH values generally range from 0 to 14:
  • A pH lower than 7 indicates an acidic solution.
  • A pH of 7 is considered neutral.
  • A pH higher than 7 indicates a basic solution.
For hypobromous acid in this specific case, the pH is 5.05, indicating that the solution is acidic. An essential point to remember is that the pH scale is logarithmic鈥攁 change of one pH unit reflects a tenfold change in hydrogen ion concentration. In real-world applications, pH values are used widely to monitor and control the acidity of substances in various industries and natural processes.

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Most popular questions from this chapter

When a solution prepared by dissolving \(4.00 \mathrm{g}\) of an unknown monoprotic acid in \(1.00 \mathrm{L}\) of water is titrated with \(0.600 \mathrm{M} \mathrm{NaOH}, 38.7 \mathrm{mL}\) of the \(\mathrm{NaOH}\) solution is needed to neutralize the acid. Determine the molarity of the acid solution. What is the molar mass of the unknown acid?

Determine the pH of the following buffer solutions. (a) \(20.0 \mathrm{mL}\) of \(0.050 \mathrm{M} \mathrm{HCN}(a q)\) is mixed with \(80.0 \mathrm{mL}\) of \(0.030 \mathrm{M} \mathrm{NaCN}(a q),\) where the \(\mathrm{p} K_{\mathrm{a}}\) of HCN is equal to 9.31. (b) \(40.0 \mathrm{mL}\) of \(0.30 \mathrm{MPIPES}\) acid is reacted with \(60.0 \mathrm{mL}\) of \(0.15 \mathrm{MPIPES}\) base.

Can an aqueous solution have a pH of zero? Explain your answer using aqueous HCl as your example.

Give two examples of a situation where you would want a buffer to have unequal amounts of the conjugate acid and the conjugate base.

Assume that you have a dilute solution of HCl \((0.10 M)\) and a concentrated solution of acetic acid \((5.0 M) .\) Which solution is more acidic? Explain.
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