Some K2SO3 and KHSO3 are dissolved in 150.0 mL of solution and the resulting pH is 7.40. Which is greater in this buffer solution, the concentration of SO32 or the concentration of HSO3? Ka (HSO3) = 1.0 × 10-7 Oconcentration of SO32- concentration of HSO3 Oboth are equal If [SO32] [HSO3] = = 0.60 M in this solution, calculate the concentration of HSO3 M
The Correct Answer and Explanation is :
To solve this, we need to evaluate the relationship between the concentrations of ( \text{SO}_3^{2-} ) and ( \text{HSO}_3^- ) in a buffer solution, given the pH and the dissociation constant ( K_a ) of ( \text{HSO}_3^- ).
Step 1: Write the equilibrium expression
The dissociation of ( \text{HSO}_3^- ) in water is:
[
\text{HSO}_3^- \rightleftharpoons \text{H}^+ + \text{SO}_3^{2-}
]
The acid dissociation constant ( K_a ) for ( \text{HSO}_3^- ) is given as ( 1.0 \times 10^{-7} ).
[
K_a = \frac{[\text{H}^+][\text{SO}_3^{2-}]}{[\text{HSO}_3^-]}
]
Step 2: Use the pH to find ([H^+])
The pH of the solution is given as 7.40. To find the concentration of hydrogen ions, we use the equation:
[
\text{pH} = -\log[\text{H}^+]
]
Rearranging to solve for ( [\text{H}^+] ):
[
[\text{H}^+] = 10^{-\text{pH}} = 10^{-7.40} \approx 4.0 \times 10^{-8} \, \text{M}
]
Step 3: Set up the equilibrium expression using the concentrations
We are given that the concentrations of ( \text{SO}_3^{2-} ) and ( \text{HSO}_3^- ) are equal, both at 0.60 M, initially.
At equilibrium, using the equation for ( K_a ), we have:
[
K_a = \frac{[\text{H}^+][\text{SO}_3^{2-}]}{[\text{HSO}_3^-]}
]
Substituting the known values:
[
1.0 \times 10^{-7} = \frac{(4.0 \times 10^{-8})(0.60)}{[\text{HSO}_3^-]}
]
Solving for ( [\text{HSO}_3^-] ):
[
[\text{HSO}_3^-] = \frac{(4.0 \times 10^{-8})(0.60)}{1.0 \times 10^{-7}} = 2.4 \times 10^{-8} \, \text{M}
]
Step 4: Conclusion
From the calculation, the concentration of ( \text{SO}_3^{2-} ) (0.60 M) is greater than the concentration of ( \text{HSO}_3^- ) (2.4 × 10⁻⁸ M). Therefore, the concentration of ( \text{SO}_3^{2-} ) is greater than the concentration of ( \text{HSO}_3^- ).
Final Answer:
The concentration of ( \text{SO}_3^{2-} ) is greater in this buffer solution.
Explanation:
- The ( \text{SO}_3^{2-} ) concentration is much higher than that of ( \text{HSO}_3^- ), as the dissociation constant for ( \text{HSO}_3^- ) is quite small (( K_a = 1.0 \times 10^{-7} )), meaning most of the ( \text{HSO}_3^- ) exists as ( \text{SO}_3^{2-} ) in the solution at equilibrium.