Part A) A solution is prepared by dissolving 0.23 mol of butanoic acid and 0.27

Question

Part A) A solution is prepared by dissolving 0.23 mol of butanoic acid and 0.27 mol of sodium butanoate in water sufficient to yield 1.00L of solution. The addition of 0.05 mol of HCl to this buffer solution causes the pH to drop slightly. The pH does not decrease drastically because the HCl reacts with the _________present in the buffer solution. The Ka of butanoic acid is 1.36x10 -3

a) H2O b)H3O+ c) butanoate ion d) butanoic acid

Part B) What is the hydronium ion concentration in a solution prepared by mixing 50.00mL of 0.11M HCN with 50.00mL of 0.043M NaCN? Assume that the volumes of the solutions are additive and that Ka=4.9x10 -10 for HCN.

Explanation / Answer

Part A)

answer : butanoate ion

CH3CH2CH2COO-   +   HCl   -----------------> CH3CH2CH2COOH

here HCl reacts with butanoate ion . which is present in buffer solution.

Part B)

pKa = 9.31

millimoles of HCN = 50 x 0.11 = 5.5

millimoles of NaCN = 50 x 0.043 = 2.15

pH = pKa + log [salt / acid]

      = 9.31 + log [2.15 / 5.5]

pH = 8.90

[H3O+] = 1.26 x 10^-9 M

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