Cyanic acid,HOCN,ionizes in water to produce the hydronium ion,H₃O⁺,and the cyanate ion,OCN^-: HOCN (aq) + H₂O (l) $\leftrightarrow$ H₃O⁺ (aq) + OCN^- (aq) K = 3.5 × 10^-⁴
If [H₃O⁺] = 7.3 × 10^-⁴ M and [OCN^-] = 7.3 × 10^-⁴ M,what is the value of the acid concentration,[HOCN]?

Question

Cyanic acid,HOCN,ionizes in water to produce the hydronium ion,H₃O⁺,and the cyanate ion,OCN^-: HOCN (aq) + H₂O (l)

\leftrightarrow

H₃O⁺ (aq) + OCN^- (aq) K = 3.5 × 10^-⁴
If [H₃O⁺] = 7.3 × 10^-⁴ M and [OCN^-] = 7.3 × 10^-⁴ M,what is the value of the acid concentration,[HOCN]?

Quizplus · Accepted Answer

Given the equilibrium constant (K) for the ionization of HOCN in water is 3.5 × 10^-4 and the concentrations of both H3O+ and OCN- are 7.3 × 10^-4 M, we can use the expression for the equilibrium constant (K = [H3O+][OCN-]/[HOCN]) to find the concentration of HOCN. Rearranging the equation to solve for [HOCN] gives [HOCN] = [H3O+][OCN-]/K. Substituting the given values, [HOCN] = (7.3 × 10^-4 M * 7.3 × 10^-4 M) / (3.5 × 10^-4) = 1.5 × 10^-3 M.

Cyanic Acid,HOCN,ionizes in Water to Produce the Hydronium Ion,H3O+,and the Cyanate

Cyanic Acid,HOCN,ionizes in Water to Produce the Hydronium Ion,H₃O⁺,and the Cyanate