What is the pOH of 0.1940 M hydrocyanic acid solution, hydrocyanic, if Ka = 4.90

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sol take help Conceptually, we know that hydrofluoric acid is a corrosive but weak acid, so it does not completely dissociate: HF ==> H+ + F- using an ICE table: HF H+ F- i: 0.065M c: -x +x +x e: (0.065 - x)M xM xM then, you use the arrange the formula as: [F-][H+] / [HF] = Ka. x^2 / (0.065 - x) = 7.2*10^-4 we can assume, that (0.065 - x) basically equals 0.065: x^2 / 0.065 = 7.2*10^-4 multiplying the denominator to the Ka value, we get: x^2 = 4.68*10^-5 x = 6.84*10^-3 this x-value is the concentration of the the H3O generated by the dissociation of HF. [H3O] = 6.84*10^-3 so, the solution's pH can be calculated: -log [H3O] = 2.16

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