HELP!A certain weak base has a Kb of 8.80 × 10-7. What concentration of this base will produce a pH of 10.11?
A certain weak base has a Kb of 8.80 × 10-7. What concentration of this base will produce a pH of 10.11?
I think:
Since the Ph is 10.11
PoH: 14-10.11=9.89
PoH=-log(OH-)=9.89
log[oh-]=-9.89
[OH-]=10^-9.89 approx: 1.28*10^-10
Kb=[BH+][OH-]/[B]
Kb=[BH+][1.28*10^-10]/[B]=8.80*10^-7
This is where I am stuck! Please help/ Explain
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Verified answer
Every step you detailed is correct. You got stuck here:
Kb=[BH+][1.28*10^-10]/[B]=8.80*10^-7
The reaction is:
B + H2O <==> BH+ + OH-
The fact that you missed is this:
[BH+] = [OH-]
Consequently,
8.80*10^-7 = [1.28*10^-10][1.28*10^-10] / [B]
and on to the concentration of B
8.80*10^-7 = (1.28*10^-10)^2 / B
B * 8.80*10^-7 = (1.28*10^-10)^2
B * 8.80*10^-7 = 1.6384*10^-20
B = 1.6384*10^-20 / 8.80*10-7
B = 1.86*10^-14
14-10.11=3.89 not 9.89