Question

Captain Kirk of the Starship Enterprise has been told by his superiors that only a chemist can be trusted with the combination of the safe containing the dilithium crystals that power the ship. The combination is the pH of solution A, described below followed by the pH of solution C. The chemist must determine the combination using only the information below (all solutions are at 25C).

Solution A is 150.0 mL of a 0.150 M solution of the weak monoprotic acid HX.

Solution B is a 0.0500 M solution of the salt NaX. It has a pH of 10.20.

Solution C is made by adding 25.0 mL of 0.200 M KOH to solution A.

What is the combination to the safe?

Answer 1

To find the pH of solution A, the precise dissociation constant of the acid HX is required as it is weak. This is obtained from solution B through the pH of the salt of the weak acid. Salt hydrolysis of salts made from weak acid and strong base will result in an unstable anionic part X^- which tends to abstract protons from the medium making it alkaline. This also explains the basic pH of NaX.

Given the concentration of NaX and the pH of that solution, we can find the K_b of X^- and thus relate it to K_a of HX. This is achieved by first finding the pOH of the solution as pOH = 14-pH = 3.8. From this, the concentration of hydroxide ions in the solution is found by the relation [OH-] = 10 - pOH = 10 - 3.8 = 1.5849 x10^-4. Now, [OH^-] = sq.rt of K_bxC_b where C_b is the concentration of base which here is 0.05. This gives [OH^-]² divided by C_b = K_b. This gives (2.5119x10^-8)/0.05 = 5.0238x10^-7. Now that K_b is known, pK_a of the acid HX is found as pKa = 14 - pK_b = 14 - 6.2989 = 7.7011.

From the pK_a of the acid, K_a is found as 10 - pK_a = 1.9902x10^-8. Now since HX is a monoprotic acid, as per Henderson-Hasselbach equation, we get K_a = [H⁺][X^-]/[HX] in which both terms in the numerators are equal due to the 1:1 ratio of H⁺ and X^-. This gives K_a as [H⁺]²/[HX] which implies 1.9902x10^-8 = [H⁺]²/[HX]. [HX] is found in no.of moles of HX by finding the no.of moles of 0.15M HX in 1mL of the solution and extending it to 150mL. Thus, we get no.of moles of HX in 150mL as (0.15/1000)x150 = 0.0225moles. Now, we get [H⁺] = sq.rt of (1.9902x10^-8)/0.0225 = 9.4049x10^-4. From this pH = -log[H⁺] = 3.0266.

Now, 25mL of 0.2M KOH contains (0.2/1000)x25 = 0.005moles of KOH. We already know that there are 0.0225moles of HX in the solution. This gives a total of 0.005moles of KX (Law of conservation of mass) and 0.0225 - 0.005 = 0.0175moles of HX remaining in the solution. Ignoring common ion effect due to absence of required parameters to take it into account, we get the pH of the solution from 0.0175moles of HX as negative logarithm of sq.rt of (1.9902x10^-8/0.0175) = 2.9721.

This gives the combination of the safe as 3.0266 and 2.9721.