## Chemistry 10th Edition

Percent ionization $\approx$ 4.9%.
1. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [HCOO^-] = x$ -$[HCOOH] = [HCOOH]_{initial} - x = 0.0751 - x$ For approximation, we consider: $[HCOOH] = 0.0751M$ 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][HCOO^-]}{ [HCOOH]}$ $Ka = 1.8 \times 10^{- 4}= \frac{x * x}{ 7.51 \times 10^{- 2}}$ $Ka = 1.8 \times 10^{- 4}= \frac{x^2}{ 7.51 \times 10^{- 2}}$ $1.352 \times 10^{- 5} = x^2$ $x = 3.677 \times 10^{- 3}$ Percent ionization: $\frac{ 3.677 \times 10^{- 3}}{ 7.51 \times 10^{- 2}} \times 100\% = 4.896\%$ %ionization < 5% : Right approximation.