# Chapter 17 - Additional Aspects of Acid-Base Equilibria - Example 17-2 - Demonstrating the Common Ion Effect: A Solution of a Weak Acid and a Salt of that Weak Acid - Page 749: Practice Example A

$[H_3O^+] = 1.2 \times 10^{-4} \space M$ $[HCOO^-] =0.150 \space M$

#### Work Step by Step

$[HCOO^-] = [NaHCOO] = 0.150 M$ 1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HCOOH ]& [ HCOO^- ]& [ H_3O^+ ]\\ Initial& 0.100 & 0.150 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.100 -x& 0.150 +x& 0 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ HCOO^- ][ H_3O^+ ]}{[ HCOOH ]}$$ $$K_a = \frac{( 0.150 + x)(x)}{[ HCOOH ]_{initial} - x}$$ 3. Assuming $0.100 \space and \space 0.150 \gt\gt x:$ $$K_a = \frac{( 0.150 )(x)}{[ HCOOH ]_{initial}}$$ $$x = \frac{K_a \times [ HCOOH ]_{initial}}{ 0.150 } = \frac{ 1.8 \times 10^{-4} \times 0.100 }{ 0.150 }$$ $x = 1.2 \times 10^{-4}$ 4. Test if the assumption was correct: $$\frac{ 1.2 \times 10^{-4} }{ 0.100 } \times 100\% = 0.12 \%$$ 5. The percent is less than 5%. Thus, it is correct to say that $x = 1.2 \times 10^{-4}$ 6. $$[H_3O^+] = x = 1.2 \times 10^{-4}$$ $[HCOO^-] = 0.150 + 1.2 \times 10^{-4} = 0.150 \space M$

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