Question

Question 26 1 pts Calculate the pH of a 0.432 M (CH3)2NH2Br solution. Remember that only the decimal places in a pH represent significant figures. Please enter your answer with three decimal places. Question 27 1 pts Calculate the pH of a 0.421 M KClO solution. Remember that only the decimal places in a pH represent significant figures. Please enter your answer with three decimal places.

Answer 1

Soln 1 ) GIVEN

conc of (CH3)2NH2Br = 0.432 M

Kb of (CH3)2NH = 5.42 x 10^-4

=> Ka = 10^-14 / Kb

= 1.845 x 10^-11

now , we know that

(CH3)2NH2Br    < ---------------------------------------> (CH3)2NH2^+    + Br-

=> (CH3)2NH2^+ + H2O <--------------------------------------> (CH3)2NH + H3O+

Ka = [(CH3)2NH][H3O^+] / [(CH3)2NH2+]

=> 1.84 x 10^-11 = x^2 / 0.432 ( since Ka is very small , 0.432 - x = 0.432)

=> x = 2.82 x 10^-6

=>[H3O^+] = 2.82 X 10^-6 M

we know that ,

pH = - log[H3O^+]

=> pH = 5.55 ANS

Soln 2) GIVEN

conc of KClO = 0.421 M

Ka of HClO = 2.95 x 10^-8

=> Kb = 10^-14 / Ka

= 3.39 x 10^-7

we know that

KClO <-------------------------------> K ^+ + ClO^-

ClO^- + H2O <-------------------------------> HClO + OH^-

Kb = [HClO][OH-] / [ClO-]

=> 3.39 x 10^-7 = x^2 / 0.421 ( since Kb is very small , 0.421 - x = 0.421)

=> x = 3.78 x 10^-4

=>[OH-] = 3.78 x 10^-4

we know that ,

pOH = -log[OH-]

= 3.42

and ,

pH + pOH = 14

=> pH = 10.58 ANS