Question

Hybrobromous acid, HBrO, is a weak acid. It's acid dissociation constant. Ka is 2.5 x 10-9

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.

Cl2 (g) + H2O ----> H+ + Cl- + 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 HOCL in the above solution is 0.065 molar (ignore any equilibrium reactions that result from the partial dissociation of HOCL)

Answer 1

a) to Lea'm: HBO ² H+ C 0 C-cd CL Ka= (Ed (cd) Tc-ca) + Obr- 0 ca 14 LULE TTE

C. concentration of HBrO = 0.14M &: degree of dissociation Ka - Co2 ( Ka=ca² 2,5x10 9 = 0.1422 & d = 1.3368104 CH+] = Cd = 0.14(1.336x164) [ht] = 1,87xos – NaBrot (aq) O el Nauh + HOB (aq.) (ag. )

log[NgoBo} = 2.0599 xl03 THOR] K= [Nach] [HOBI (Naoße] [40] NOTE: Numerical value of ea'm constant Can be calculated Becaux conc. terms are not geven c) 40 mL of 0.14M HBrO [NaoBr] = 0.99527 THobr] (Naobr] =0.99527 [hober] 0.99527 (0.2M) = 0.19905 Home x o 14 mol = 5.6 mmol of Horo mass - Mobiy xoay poss x Volume 5 ml of 0.56M Na oH 5my xo 156 mol z 208 mmol of Nach = 0.199105 X 118.99 x 50 xh of you mass = 1.1834g 1-1834 g of NaoBe must be added to the solution that pH becomes 8.60 HBRO + NaOH Na o Bo 1720 . 2.8 mmol of Nauh will react with Sot me 208 mmol of HBO froduing 2.8 mmol of Naobr [Naobr] = n NauBr = 2.8 mmol = 0.0622M Volume (4015)he [. Bro] = (5.6-28 ) mmol = 0.0622M Mots) mL pH = pka + log [salt] Noober CAuid] HBO pH = 8.602 tlog (0.062) (pka = -logka = -log(2.5x59) = 8.602 [PH = 8.62 (": log (1) = 0) d) To have a buffer of pH = 8.60 8.60= 8.602 tlog [NapBo] THOB