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_aKa.
The molar mass, mm, of formic acid is 5050 gmol^(-1)gmol−1, so the number of moles is given by n=m/M = 23/50 = 0.46n=mM=2350=0.46 molmol.
Dissolving 0.460.46 molmol of formic acid in 1010 LL of water yields to a concentration of C=n/V = 0.46/10 = 0.046C=nV=0.4610=0.046 molmol.
In this case, formic acid dissociates as follows:
HCOOH \rightleftharpoons H^+ + HCOO^-HCOOH⇌H++HCOO−
The expression for K_aKa is as follows:
K_a = ([H^+][HCOO^-])/([HCOOH])Ka=[H+][HCOO−][HCOOH]
(we can ignore [H_2O][H2O] because it is effectively constant)
We are told that [H^+][H+] = 3 \times 10^-33×10−3 molL^-1molL−1
(molL^-1molL−1 is a more technically-correct way than MM of writing a concentration in SI units)
Looking at the equation, each mole of formic acid that dissociates yields one mole of H^+H+ and one mole of HCOO^-HCOO−. Substituting these values into the K_aKa expression yields:
K_a = ([H^+][HCOO^-])/([HCOOH])Ka=[H+][HCOO−][HCOOH] = ([3 \times 10^-3][3 \times 10^-3])/([0.046 - (3 \times 10^-3)]) = 2.1 \times 10^(-4)=[3×10−3][3×10−3][0.046−(3×10−3)]=2.1×10−4
This is a unitless constant.
