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Question

1(a). The pH of a 0.0060 M solution of butanoic acid, HC4H7O2,is 3.52, but the pH of a 0.0060 M solution of perchloric acid is 2.22. Why is the pH of the butanoic acid solution different from the pH of the perchloric acid solution? Explain. 

(b) Find the concentration of perchloric acid in a perchloric acid solution having a pH of 3.52. 

2. You are asked to characterize the newly discovered acid periodic acid (HIH4O6, M= 227.94g/mol). One of your tasks is to establish whether it is a strong or weak acid. Hence, you prepare a solution by dissolving 0.457g of periodicacid in enough water to make 40.0 mL solution. The pH of this solution is 1.96. Is periodic acid a strong or weak acid? Show by a suitable calculation.

Answer 1

I will explain and guide you, but you should read about weak acids, strong acids, pH and Ka values and what they mean.

1(a). The pH of butanoic acid and perchloric acid at the same concentration is different because the 2 acids ionize to different extents. This will be given by the respective Ka values for the different acids. In this case, the lower pH of the perchloric acid solution indicates that it ionizes to a greater extent than does the butanoic acid, so it would have a higher Ka. It will provide more H+ ions per mole than will butanoic acid, and thus have a lower pH.

(b). pH = 3.52

[H+] = 1x10^-3.52 = 3.02x10^-4 M

HClO4 ==> H+ + ClO4

Ka = [H+][ClO4-]/[HClO4] = (2.02x10^-4)(2.02x10^-4) / [HClO4]. Look up the Ka for perchloric acid (HClO4), plug it into this equation and solve for [HClO4]. If you are in a more advanced class, you will know that perchloric acid (HClO4) is a strong acid, and it will not have a Ka. Thus, [HClO4] = 3.02x10^-4 M as it will ionize 100% producing 3.02x10^-4 M H+ and 3.02x10^-4 M ClO4^-.

2 . Find molarity by dividing mass of 0.457 g by molar mass of 227.94 g. Divide that by the liters of solution (0.04 L). This gives you molarity of the solution (M).

Represent this acid as HA. Then you have HA ==> H+ + A- and you can find [H+] and [A-] from the given pH. [H+] = [A-] = 1x10^-1.96 = 1.1x10^-2 M.

Compare this value to the value you calculated for the molarity of the acid solution. If it is the same, or very close, then you would have a strong acid. If it is considerably different, then the acid is weak, as it didn't ionize 100%