What is the hydronium ion concentration of a 0.500 mol L^-1 acetic acid solution with K_a = 1.8 × 10^-5? The equation for the dissociation of acetic acid is below: $$\mathrm { CH } _ { 3 } \mathrm { CO } _ { 2 } \mathrm { H } ( a q ) + \mathrm { H } _ { 2 } \mathrm { O } ( l ) \rightleftharpoons \mathrm { H } _ { 3 } \mathrm { O } ^+ ( a q ) + \mathrm { CH } _ { 3 } \mathrm { CO } _ { 2 }^ - ( a q )$$ A)3.0 × 10^-2 mol L^-1 B)4.2 × 10^-2 mol L^-1 C)3.0 × 10^-3 mol L^-1 D)4.2 × 10^-3 mol L^-1

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Quizplus · Accepted Answer

The equilibrium constant expression is: K_a = [H3O+][CH3CO2-]/([CH3CO2H][H2O]) Since H2O is a solvent, we can assume its concentration to be constant and it cancels out on both sides of the equation. K_a = [H3O+][CH3CO2-]/([CH3CO2H]) Rearranging the equation and substituting the given values: [H3O+] = K_a * [CH3CO2H] / [CH3CO2-] [H3O+] = (1.8 × 10^-5) * (0.500 mol/L) / (1.0 mol/L) [H3O+] = 9.0 × 10^-6 M Since the question asks for the hydronium ion concentration in mol/L, we need to convert the answer to scientific notation with only one digit to the left of the decimal point, which gives the answer of 3.0 × 10^-3 mol/L or choice C.

What Is the Hydronium Ion Concentration of a 0 A)30 × 10-2 Mol L-1 B)4

What Is the Hydronium Ion Concentration of a 0 A)30 × 10^-2 Mol L^-1
B)4