Asked by Margaret
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).
Answers
Answered by
Devron
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
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
Answered by
Anonymous
9.5x10^-16
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