This is what I want to do but I am not sure.
HONH2 + H2O---> OH + H2ONH2
14-pH=pOH=3.98
10^(-3.98)=OH concentration=1.05 x 10^-4 M
1.1 x 10^-8=Kb=[1.05 x 10^-4 M][1.05 x 10^-4 M]/[x-1.05 x 10^-4 M]
Solve for x, 5% rule allows us to eliminate 1.05 x 10^-4 M from the denominator.
([1.05 x 10^-4 M][1.05 x 10^-4 M]/1.1 x 10^-8)=x
x =molarity of HONH2
Molarity of HONH2*(255.5 x 10^-3L)= moles of HONH2
moles of HONH2*(47.013 g of HONH2/mol)= mass of HONH2
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).
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