Consider a 2.5 M solution of sodium butanoate at 25°C а Given: Kal CH3CH2CH2COOH) = 1.5 x 10-5, and Kw = 1.0 X 10-14 Ans

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Consider a 2.5 M solution of sodium butanoate at 25°C а Given: Kal CH3CH2CH2COOH) = 1.5 x 10-5, and Kw = 1.0 X 10-14 Answer the following questions (note any instructions provided in a different colourt); The value of Kb for the butanoate ion is = (The answer must be expressed in exponential notation, og 0.0010 is the same as 1.02-3; no spaces anywhere!) Use the variable x to indicate the changes in concentration when equilbrium is reached The expression involving x that shows the concentration of the butanoate ion at equilibrium is (No spaces anywhere!) The calculated value of x is (You may make use of an approximation if you wish) (Express your answer in exponential notation to two sig. figs) The value of the pOH is (Express your answer as a dealmat to the correct number of sigfign) The value of the pH is then (Express your answer as a decimal to the correct number of sig figs.) The percentage ionisation of the original propanaote ion is (Express your answer as a decimal to the correct number of sig, fig.)