Calculate the mass of HONH2 required to dissolve in enough water to make 255.5 mL of solution having a pH of 10.02 (Kb = 1.110−8).

Answer 1

pOH=14-pH=14-10.02=3.98

pOH=-log[OH-]

[OH-]=10^(-3.98)=1.05 x 10^-4 M

HONH2 + H2O ----.> OH- + HONH3

Kb=1.1 x 10−8=[1.05 x 10^-4 M][1.05 x 10^-4 M]/(x-1.05 x 10^-4 M)

5% rule states that we can ignore -1.05 x 10^-4 M.

Solving for x,

Kb=1.1 x 10−8=[1.05 x 10^-4 M][1.05 x 10^-4 M]/x

x=1.21 x 10^-16 M

Molarity=1.21 x 10^-16 M=moles/volume (L)

Solving for moles,

1.21 x 10^-16 M *(255.5 x 10^-3L)= moles of HONH2

moles of HONH2 *(33.03g of HONH2/mole)= mass of HONH2

Answer 2

9.5x10^-16