Question #bd092

1 Answer
Mar 22, 2017

Answer:

#K_a = ([H^+][HCOO^-])/([HCOOH])## = ([3 \times 10^-3][3 \times 10^-3])/([0.046 - (3 \times 10^-3)]) = 2.1 \times 10^(-4)#

Explanation:

I'm afraid I don't know what 'ICE BOX' means in this context: I assume it's a mnemonic device intended to help you remember how to calculate a #K_a#.

The molar mass, #m#, of formic acid is #50# #gmol^(-1)#, so the number of moles is given by #n=m/M = 23/50 = 0.46# #mol#.

Dissolving #0.46# #mol# of formic acid in #10# #L# of water yields to a concentration of #C=n/V = 0.46/10 = 0.046# #mol#.

In this case, formic acid dissociates as follows:

#HCOOH \rightleftharpoons H^+ + HCOO^-#

The expression for #K_a# is as follows:

#K_a = ([H^+][HCOO^-])/([HCOOH])#

(we can ignore #[H_2O]# because it is effectively constant)

We are told that #[H^+]# = #3 \times 10^-3# #molL^-1#

(#molL^-1# is a more technically-correct way than #M# of writing a concentration in SI units)

Looking at the equation, each mole of formic acid that dissociates yields one mole of #H^+# and one mole of #HCOO^-#. Substituting these values into the #K_a# expression yields:

#K_a = ([H^+][HCOO^-])/([HCOOH])## = ([3 \times 10^-3][3 \times 10^-3])/([0.046 - (3 \times 10^-3)]) = 2.1 \times 10^(-4)#

This is a unitless constant.