Question

S> Question 7 (2 points) What is the pH of a 0.174 M aqueous solution of hydroxylamine, HONH2? The Ky for methylamine is 1.1x10-8. You do not need to solve the quadric equation for this problem. Express your answer to two decimal places. Your Answer: Answer

Answer 1

HONH2 dissociates as:

HONH2 +H2O     ----->     HONH3+   +   OH-
0.174                   0         0
0.174-x                 x         x


Kb = [HONH3+][OH-]/[HONH2]
Kb = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes

Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((1.1*10^-8)*0.174) = 4.375*10^-5

since c is much greater than x, our assumption is correct
so, x = 4.375*10^-5 M



So, [OH-] = x = 4.375*10^-5 M


use:
pOH = -log [OH-]
= -log (4.375*10^-5)
= 4.359


use:
PH = 14 - pOH
= 14 - 4.359
= 9.641
Answer: 9.64