# Question #bd092

Mar 22, 2017

${K}_{a} = \frac{\left[{H}^{+}\right] \left[H C O {O}^{-}\right]}{\left[H C O O H\right]}$$= \frac{\left[3 \setminus \times {10}^{-} 3\right] \left[3 \setminus \times {10}^{-} 3\right]}{\left[0.046 - \left(3 \setminus \times {10}^{-} 3\right)\right]} = 2.1 \setminus \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 $50$ $g m o {l}^{- 1}$, so the number of moles is given by $n = \frac{m}{M} = \frac{23}{50} = 0.46$ $m o l$.

Dissolving $0.46$ $m o l$ of formic acid in $10$ $L$ of water yields to a concentration of $C = \frac{n}{V} = \frac{0.46}{10} = 0.046$ $m o l$.

In this case, formic acid dissociates as follows:

$H C O O H \setminus r i g h t \le f t h a r p \infty n s {H}^{+} + H C O {O}^{-}$

The expression for ${K}_{a}$ is as follows:

${K}_{a} = \frac{\left[{H}^{+}\right] \left[H C O {O}^{-}\right]}{\left[H C O O H\right]}$

(we can ignore $\left[{H}_{2} O\right]$ because it is effectively constant)

We are told that $\left[{H}^{+}\right]$ = $3 \setminus \times {10}^{-} 3$ $m o l {L}^{-} 1$

($m o l {L}^{-} 1$ 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 $H C O {O}^{-}$. Substituting these values into the ${K}_{a}$ expression yields:

${K}_{a} = \frac{\left[{H}^{+}\right] \left[H C O {O}^{-}\right]}{\left[H C O O H\right]}$$= \frac{\left[3 \setminus \times {10}^{-} 3\right] \left[3 \setminus \times {10}^{-} 3\right]}{\left[0.046 - \left(3 \setminus \times {10}^{-} 3\right)\right]} = 2.1 \setminus \times {10}^{- 4}$

This is a unitless constant.