Directions: Questions 4-7 are short free-response questions that require about 9 minutes each to answer and are worth 4 points each. Write your response in the space provided following each question. Examples and equations may be included in your responses where appropriate. For calculations, clearly show the method used and the steps involved in arriving at your answers. You must show your work to receive credit for your answer. Pay attention to significant figures. Hyprobromous acid, HBrO, is a weak monoprotic acid with a \(K_{\mathrm{a}}\) value of \(2.0 \times 10^{-9} \mathrm{at} 25^{\circ} \mathrm{C} .\) (a) Write out the equilibrium reaction of hyprobromous acid with water, identifying any conjugate acid/based pairs present. (b) (i) What would be the percent dissociation of a 0.50 M solution of hyprobromous acid? (ii) If the 0.50 M solution were diluted, what would happen to the percent dissociation of the HBrO? Why?

Step by step solution

01

Identify the reaction

Hyprobromous acid reacts with water by losing a proton to form hydronium ion and bromite ion in an aqueous solution. The reaction can be written as: \[ HBrO + H_2O ⇌ H_3O^+ + BrO^- \]. The conjugate acid/base pairs are \(HBrO / BrO^-\) and \(H_3O^+ / H_2O\).

02

Calculate the percent dissociation

The equilibrium constant expression is: \[ Ka = \frac{[H_3O^+][BrO^-]}{[HBrO]} \] Initially, only HBrO is present, so at equilibrium: \[2.0 \times 10^{-9} = \frac{x^2}{0.50 - x} \] where x is the number of moles of hydronium and bromite ions at equilibrium. Without doing any math you realize, that due to the very small equilibrium constant, the value of x is expected to be very much smaller than 0.50, meaning it can be neglected in the denominator: \[2.0 \times 10^{-9} = \frac{x^2}{0.50} \] From which we can solve for x. \(x = \sqrt{2.0 \times 10^{-9} \times 0.50} = 3.16 \times 10^{-5} M\), which represents the molar concentration of hydronium ions or the dissociated acid. \[ percent dissociation = \left(\frac{x}{0.50}\right) \times 100% = \left(\frac{3.16 \times 10^{-5}}{0.50}\right) \times 100% = 6.32 \times 10^{-3}% \] The percent dissociation is 6.32 \times 10^{-3} percent.

03

Effect of Dilution on Percent Dissociation

If the HBrO solution were diluted, the percent dissociation would increase. This is because of LaChatelier’s Principle, which states that a system at equilibrium will shift to counteract a stress placed upon it. When you add more water (diluting the solution), the system shifts to produce more ions to increase the ion concentration, thereby increasing the percent dissociation.

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