A 0.645 M solution of a weak base (B:) is made. The solution has a pH of 11.99. Calculate the Kb of this base. Report your answer in scientific notation with 3 sig figs.
Consider reaction, B (aq) + H2O (l) BH + (aq) + OH - (aq)
Equilibrium constant for above reaction is K b = [ BH + ] [ OH - ] / [ B ]
According to above reaction, [ BH + ] = [ OH - ]
We can calculate [ OH - ] from pH.
We have relation, pH = - log [ H3O + ]
[ H3O + ]= 10 -pH
[ H3O + ] = 10 - 11.99
[ H3O + ]=10.23 10 - 12 M
We have relation, [ H3O + ] [ OH - ] = 1.00 10 -14
[ OH - ] = 1.00 10 -14 / [ H3O + ]
[ OH - ] = 1.00 10 -14 / 10.23 10 - 12
[ OH - ] = 9.77 10 -03 M
Let's use ICE table.
Concentration (M) | B (aq) BH + (aq) + OH - (aq) | ||
I | 0.645 | ||
C | - 9.77 10 -03 | + 9.77 10 -03 | + 9.77 10 -03 |
E | 0.645 - 9.77 10 -03 | 9.77 10 -03 | 9.77 10 -03 |
We have, K b = [ BH + ] [ OH - ] / [ B ]
K b = ( 9.77 10 -03 ) ( 9.77 10 -03 ) / 0.645 - 9.77 10 -03
K b = 1.5026 10 -04
ANSWER : K b = 1.50 10 -04
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