How many milligram of sodium acetate should be added in 500 ml of 0.2 M acetic acid solution in orde — Ionic Equilibrium Chemistry Question
Question
How many milligram of sodium acetate should be added in 500 ml of 0.2 M acetic acid solution in order to make the $[\text{H}^+]$ in the solution, $4 \times 10^{-4}\text{ M}$. $K_a$ of $\text{CH}_3\text{COOH} = 1.8 \times 10^{-5}$.
💡 Solution & Explanation
This forms an acidic buffer. From the buffer equation, $[\text{H}^+] = K_a \times \frac{[\text{Acid}]}{[\text{Salt}]} \implies 4 \times 10^{-4} = 1.8 \times 10^{-5} \times \frac{0.2}{[\text{Salt}]}$. Solving for $[\text{Salt}]$ yields $\frac{3.6 \times 10^{-6}}{4 \times 10^{-4}} = 0.009\text{ M}$. The moles of sodium acetate in $500\text{ ml}$ is $0.009 \times 0.5 = 0.0045\text{ moles}$. The molar mass of $\text{CH}_3\text{COONa}$ is $82\text{ g/mol}$. The required mass is $0.0045 \times 82 = 0.369\text{ g} = 369\text{ mg}$. Formatted to four digits, this is 0369.