Asked by Shawna Williams on Mar 10, 2024
Verified
What is the K of the reaction HC2H3O2 ⇌ H+ + C2H3O2- , if [HC2H3O2] = 0.15 M and the pH = 3.0?
A) 1.00 × 10-6
B) 6.67 × 10-6
C) 3.34 × 10-6
D) 1.50 × 10-7
pH
A scale used to specify the acidity or basicity of an aqueous solution, ranging from 0 to 14.
[HCO3-]
[HCO3-] is the chemical notation for the bicarbonate ion, a key intermediate in the carbon dioxide transport and pH balance mechanisms in living organisms.
- Acquire knowledge on the notion of chemical equilibrium and the technique for computing the equilibrium constant (K).
- Relate the pH of a solution to the concentrations of reactants and products in acid-base equilibria.
Verified Answer
MA
Maria ArtisMar 10, 2024
Final Answer :
B
Explanation :
The equilibrium expression for the reaction is K=[H+][S−][CS(OH)3−][HCS(OH)3].K = \frac{[H^+][S^-][CS(OH)_3^-]}{[HCS(OH)_3]}.K=[HCS(OH)3][H+][S−][CS(OH)3−]. Given that [HCS(OH)3]=0.15 M[HCS(OH)_3] = 0.15\,M[HCS(OH)3]=0.15M and the pH is 3.0 (which means [H+]=10−3 M[H^+] = 10^{-3}\,M[H+]=10−3M ), and assuming the reaction goes to completion, [S−]=[CS(OH)3−]=10−3 M.[S^-] = [CS(OH)_3^-] = 10^{-3}\,M.[S−]=[CS(OH)3−]=10−3M. Thus, K=(10−3)(10−3)(10−3)0.15=6.67×10−6.K = \frac{(10^{-3})(10^{-3})(10^{-3})}{0.15} = 6.67 \times 10^{-6}.K=0.15(10−3)(10−3)(10−3)=6.67×10−6.
Learning Objectives
- Acquire knowledge on the notion of chemical equilibrium and the technique for computing the equilibrium constant (K).
- Relate the pH of a solution to the concentrations of reactants and products in acid-base equilibria.