a monoprotic organic acid with a Ka of 6.7*10^-4 is 35% ionized when 100g of it is dissolved in 1L. What is the formula weight of the acid? (need step wise solution)
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Calculate the molar mass of a monoprotic organic acid whose Ka is 6.7 x 10 -4 M and 3.5% dissociated when 100 g is dissolved in 1.000 L of solution. The ...
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Ans Ka = Ca^2/1-a 6.7 x 10^-4 = C [0.035 x 0.035]/1-0.035 = C [0.001225]/0.965 6.7 x 10^-4 = C [0.00126943] C = 0.527795 = 100/M M = 189.46 g/mol
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