## Chemistry: Molecular Approach (4th Edition)

1. Calculate the molar mass: $HCHO_2$ : ( 1.008 $\times$ 2 )+ ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 46.03 g/mol 2. Use the information as conversion factors to find the molarity of this solution: $$\frac{ 1.35 g \space HCHO_2 }{100 \space g \space solution} \times \frac{1 \space mol \space HCHO_2 }{ 46.03 \space g \space HCHO_2 } \times \frac{ 1.01 \space g \space solution}{1 \space mL \space solution} \times \frac{1000 \space mL}{1 \space L} = 0.296 \space M$$ 1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HCHO_2 ]& [ CH{O_2}^- ]& [ H_3O^+ ]\\ Initial& 0.296 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.296 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ CH{O_2}^- ][ H_3O^+ ]}{[ HCHO_2 ]}$$ $$K_a = \frac{(x)(x)}{[ HCHO_2 ]_{initial} - x}$$ 3. Assuming $0.296 \gt\gt x:$ $$K_a = \frac{x^2}{[ HCHO_2 ]_{initial}}$$ $$x = \sqrt{K_a \times [ HCHO_2 ]_{initial}} = \sqrt{ 1.8 \times 10^{-4} \times 0.296 }$$ $x = 7.3 \times 10^{-3}$ 4. Test if the assumption was correct: $$\frac{ 7.3 \times 10^{-3} }{ 0.296 } \times 100\% = 2.5 \%$$ 5. The percent is less than 5%. Thus, it is correct to say that $x = 7.3 x 10^{-3}$ 6. $$[H_3O^+] = x = 7.3 \times 10^{-3}$$ 7. Calculate the pH: $$pH = -log[H_3O^+] = -log( 7.3 \times 10^{-3} ) = 2.14$$