Question

SHORT ANSWER: 11. Hypobromous acid, HBO pobromous acid. HR is a weak acid. Its acid dissociation constant. K is 2.5 x 10 (20 points) a) Calculate the [H") of a 0.14-molar solution of HBrO. b) Write the correctly balanced net ionic equation for the reaction that occurs NaBro is dissolved in water and calculate the numerical value of the equilibrium constant for this reaction. c) Calculate the pH of a solution made by combining 40.0 milliliters of 0.14-molar HBrO and 5.0 milliliters of 0.56-molar NaOH. d) How many grams of solid NaBrO must be added to 50.0 milliliters of 0.200-molar HBrO to obtain a buffer solution that has a pH of 8.60? Assume that the addition of the solid NaBrO results in a negligible change in volume. e) Household bleach is made by dissolving chlorine gas in water, as represented below. Cla(g) + H2O → H+CI+HOCl(aq) Note that this reaction is NOT an equilibrium reaction. It does go to completion Calculate the pH of such a solution if the concentration of HOCI in the above solution is 0.065 molar. (Ignore any equilibrium reactions that result from the partial dissociation of HOCI). Use serap paper provided to show your work and then write the complete answers here: REMEMBER YOU UNITS!!! a) [H] = 1.87*10-5 2.5*10-9a = 1.87*10 HBrO caght H₂ Olag H₂O (aq) + BrO cag) x = 4.0*10-6 pH = 12.8 1 pH = 1.2

Answer 1

11)

a)         HBrO(aq) + H2O(l) <-----> H3O+(aq) + BrO-(aq)

Initial     0.14 M                      -         -

change        -x                         +x         +x

equil      0.14-x                        x         x

Ka = [H3O+][BrO-]/[HBrO]

2.5*10^-9 = (x*x)/(0.14-x)

x = 1.87*10^-5

at equilibrium,

[H3O+] = x = 1.87*10^-5 M

pH = -log[H3O+]

pH = -lOg(1.87*10^-5)

pH = 4.73

b)   BrO-(aq) + H2O(l) <----> HBrO(aq) + OH-(aq)

    Kh = Kw/Ka

    Kh = equilibrium constant = (1*10^-14)/(2.5*10^-9) = 4*10^-6

    Kh = 4*10^-6

c) No of mol of HBrO reacted = 40*0.14 = 5.6 mmol

   No of mol of NaOH added   = 5*0.56 = 2.8 mmol

pH of acidic buffer = pka + log(NaBrO/HBrO)

      pka of HBrO = -logKa = -log(2.5*10^-9) = 8.6

pH = 8.6+log(2.8/(5.6-2.8))

pH = 8.6

d)
No of mol of HBrO reacted = 50*0.2 = 10 mmol

No of mol of NaBrO added   = ? mmol

pH of acidic buffer = pka + log(NaBrO/HBrO)

pka of HBrO = -logKa = -log(2.5*10^-9) = 8.6

8.6 = 8.6+log(x/(10))
  
x = 10 mmol

mass of NaBrO added   = 10*10^-3*103 = 1.03 g

e) from equation,

      1 mol H+ = 1 mol Cl- = 1 mol HOCl

concentration f HOCl = 0.065 M

concentration f H+ = 0.065 M

pH = -log[H+]

   = -log0.065

   = 1.19