Captain Kirk of the Starship E
Captain Kirk of the Starship Enterprise has been told by hissuperiors that only a chemist can be trusted with the combinationof the safe containing the dilithium crystals that power the ship.The combination is the pH of solution A, described below followedby the pH of solution C. The chemist must determine the combinationusing only the information below (all solutions are at 25C).
Solution A is 150.0 mL of a 0.150 M solution of the weakmonoprotic acid HX.
Solution B is a 0.0500 M solution of the salt NaX. It has a pHof 10.20.
Solution C is made by adding 25.0 mL of 0.200 M KOH to solutionA.
What is the combination to the safe?
Answer:
To find the pH of solution A, the precise dissociation constantof the acid HX is required as it is weak. This is obtained fromsolution B through the pH of the salt of the weak acid. Salthydrolysis of salts made from weak acid and strong base will resultin an unstable anionic part X– which tends to abstractprotons from the medium making it alkaline. This also explains thebasic pH of NaX.
Given the concentration of NaX and the pH of that solution, wecan find the Kb of X– and thus relate it toKa of HX. This is achieved by first finding the pOH ofthe solution as pOH = 14-pH = 3.8. From this, the concentration ofhydroxide ions in the solution is found by the relation [OH-] = 10- pOH = 10 – 3.8 = 1.5849 x10-4. Now, [OH–] =sq.rt of KbxCb where Cb is theconcentration of base which here is 0.05. This gives[OH–]2 divided by Cb =Kb. This gives (2.5119×10-8)/0.05 =5.0238×10-7. Now that Kb is known,pKa of the acid HX is found as pKa = 14 – pKb= 14 – 6.2989 = 7.7011.
From the pKa of the acid, Ka is found as10 – pKa = 1.9902×10-8. Now since HX is amonoprotic acid, as per Henderson-Hasselbach equation, we getKa = [H+][X–]/[HX] in which bothterms in the numerators are equal due to the 1:1 ratio ofH+ and X–. This gives Ka as[H+]2/[HX] which implies1.9902×10-8 = [H+]2/[HX]. [HX] isfound in no.of moles of HX by finding the no.of moles of 0.15M HXin 1mL of the solution and extending it to 150mL. Thus, we getno.of moles of HX in 150mL as (0.15/1000)x150 = 0.0225moles. Now,we get [H+] = sq.rt of (1.9902×10-8)/0.0225 =9.4049×10-4. From this pH = -log[H+] =3.0266.
Now, 25mL of 0.2M KOH contains (0.2/1000)x25 = 0.005moles ofKOH. We already know that there are 0.0225moles of HX in thesolution. This gives a total of 0.005moles of KX (Law ofconservation of mass) and 0.0225 – 0.005 = 0.0175moles of HXremaining in the solution. Ignoring common ion effect due toabsence of required parameters to take it into account, we get thepH of the solution from 0.0175moles of HX as negative logarithm ofsq.rt of (1.9902×10-8/0.0175) =2.9721.
This gives the combination of the safe as 3.0266 and 2.9721.
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