What is the Ka for HNO2 if its starting concentration of 0.14 M decreased to 0.11 M by the time the reaction species reached equilibrium?

1 Answer
Ernest Z. · Truong-Son N.
Jun 19, 2017

Ka=0.008.

Explanation:

Let's pretend we are trying to solve this question using an ICE table. We would write

mmmmmmHNO2+H2O⇌H3O++NO-2
I/mol⋅L-1:m0.14mmmmmmml0mmmll0
C/mol⋅L-1:m-xmmmmmmml+xmmm+x
E/mol⋅L-1:l0.14-xmmmmmmllxmmmllx

In this case, we are told that the equilibrium concentration of the acid, i.e., [HNO2]=0.11 mol/L.

In other words,

0.14−x=0.11

x=0.14−0.11=0.03

If we insert this number into the ICE table, we get

mmmmmmHNO2+H2O⇌H3O++NO-2
I/mol⋅L-1:m0.14mmmmmmml0mmmll0
C/mol⋅L-1:l-0.03mmmmmm+0.03ml+0.03
E/mol⋅L-1:ll0.11mmmmmmm0.03mml0.03

We can now insert these numbers into the Ka expression.

Ka=[H3O+][NO-2][HNO2]=0.03×0.030.11=0.008

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