Use the References to access important values if needed for

Use the References to access important values if needed for this question. Calculate the pH of a 0.388 M aqueous solution of isoquinoline (C9H7N, Kb = 2.5x10-9). pH =

Solution

1) Let the base C9H7N be abbrevated as B  

we have base dissociation eq B (aq) + H2O (l) <---> BH+ (aq) + OH- (aq)

at equilibrium [B] = 0.388-X , [BH+] =[OH-] = X

Kb = [ BH+] [OH-] /[B]

2.5 x 10^ -9 = X^2 / ( 0.388-X) ( we apprxoimate 0.388 as 0.388 since Kb is small we get X small)

hence 2.5 x 10^ -9 = X^2 / ( 0.388)

X = [OH-] = 3 x 10^ -5

pOH = -llog ( 3 x 10^ -5) = 4.5

pH = 14 - 4.5 = 9.5

2) we do in same method as above , let base C5H11N be denoated as B

Kb = [BH+] [OH-] /[B]

1.3 x 10^ -3 =X^2 / ( 0.0851-X) ( here Kb is not so small , hence we dont take aproximation)

0.0013 ( 0.0851-X) = X^2

X^2 + 0.0013X -0.00011 = 0

X = [OH-] = 0.00986

pOH = -log ( 0.00986) = 2

pH = 14 - 2 = 12