1) An solution of an equimolar mixture of propanoic acid (propionic acid), CH3CH

Question

1) An solution of an equimolar mixture of propanoic acid (propionic acid), CH3CH2COOH, and its conjugate base has a pH of 4.87. Draw the structural formula that predominates at pH 7.00. Draw this structural formula with any appropriate formal charges and hydrogen atoms.

2)A weak acid (HA) has a pKa of 4.974. If a solution of this acid has a pH of 4.191, what percentage of the acid is not ionized?

3) Calculate the pH of a buffer solution obtained by dissolving 15.0 g of KH2PO4 (MM = 136.1 g/mol) and 31.0 g of Na2HPO4 (142.0 g/mol) in water and then diluting to 1.00 L.

Explanation / Answer

(a)Ka= 10^(-4.974) = 1.06*10^-5

(b)Ka= [H+][A-]/[HA] and you know [A-] = [H+]   solve for A- and H+ concentrations by taking the inverse log of the pH = 10^(-4.191) =6.44*10^-5

Now you have from the equation (b)

1.06*10^-5 = [6.44*10^-5][6.44*10^-5] / [HA]   Next you simpily solve for [HA] which should be 3.91*10^-4 M

Now that you have [HA] and the [H+] you are able to tell how much has ionized from HA to H+ by finding the percent

[HA] / [HA] +[H+] *100    yours should be the following...   (3.91*10^-4/ 3.91*10^-4 + 6.44*10^-5) *100 = 85.86%

Please check math before entering score

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