Acid - Base Challenge Problem A student is given 3 beakers: Beaker 1- 50.0 ml of

Question

Acid - Base Challenge Problem

A student is given 3 beakers:

            Beaker 1- 50.0 ml of a solution produced by dissolving 6.00 grams of a weak

                             monoprotic acid ,HX, in enough water to produce 1 liter of solution.

                             The empirical formula of HX is CH2O. The solution contains 3 drops

                             of phenolphthalein.

            Beaker 2- A 0.07M solution of the salt NaX. It has a pH of 8.8

            Beaker 3 – 50.0 ml of 0.250M KOH

The contents of beaker 3 is added drop-wise to beaker 1 until a pink color appears and remains for 30 seconds. This takes exactly 20.0 ml of the beaker 3 solution. Identify X and calculate the pH of beaker 1 after the addition of the 20 ml.

Explanation / Answer

Answer: According to the given informations in the question, Here Based on the empirical formula, the weak acid is most likely to be acetic acid, CH3COOH (C2H4O2).

Weak acid HX has a concentration of 6.00 grams per liter. 50 mL of the solution contains 0.300g of the weak acid.
50.0 mL x (6.00g / 1000 mL) = 0.300g HX

Now, A salt NaX contains the anion of the weak acid. A pH of 8.8 is basic because the anion hydrolyzes to produce OH- ions. NaX, being a sodium salt, completely ionizes.

X- + HOH <==> HX + OH- ............ Kb = Kw / Ka

pH = 8.80

pOH = 5.20 (at 25C)

OH- = 10-PoH = 10-5.20 = 6.31x10^-6M

Kb = [HX][OH-] / [X-] = Kw/Ka

Ka = Kw[X-] / ([HX][OH-])

Ka = 1.00x10-14 (0.07) / (6.31x10-6)2

Ka = 1.76x10-5

This is the Ka of acetic acid, CH3COOH, confirming the identity of HX.

Hence it is all about the given question . Thank you :)

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