Question

When dealing with buffer systems, we often use the acid reaction to determine the equilibrium and set up the ICE table. The most common experimental method used to study acid-base systems is titration, which studies the stoichiometric addition of the acid to a base solution or the base to an acid solution to determine the value of K, or K, The value of the pK, or pK, is essential to the understanding of the buffer system, where pK, = -log K, and pK, = - log K The Henderson-Hasselbach equation gives the relationship of the pH to the pK,, This equation is derived from the equilibrium expression, K, by rearranging and taking the log [H3O-A [HAJ [HA where [H,0-1]= K. and HA -log[H30-1--log (Ka) or pH pk+ log - log () The above concentrations are equilibrium concentrations of the acid and conjugate base; however, since have K, values of less than 10, the dissociation of the acid is small and we can assume that weak ac the change is negligible. The equilibrium concentrations are therefore assumed to be equal to the initial concentration. These concentrations also tend to be high (1 M) to increase the buffer capacity which makes this practice a good estimation. 1. Determine the pH of a buffer solution of 0.25 M sodium dihydrogen phosphate, NaH,PO, and 0.25 M disodium monohydrogen phosphate, Na,HPO The K, values for pho sphoric acid, H,PO, are Kg 7.1 x 10 K, = 6.3 x 10; and Kg = 4.2 x 1011 a. Write the chemical reaction of the acid undergoing dissociation to form the conjugate base. Is the sodium ion necessary for the chemical reaction? Which of the above compounds would act as the acid and which would act as the conjugate base? b. What is the K, expression for the reaction? Which of the above K, values are needed for the reaction? Determine the value of the pK? C.

Answer 1

(1) (a) Nahpoy & Naghoy both are strong eletrolytes & fully dissociate as. Natapoy Nat + H₂ pour ...6). NGAPO = 2Nat & Hp07-... (ii) Here, Hapoy act as weak acidf #po in conjugate bare of weak oid Hapoy. . here, [Hapoy [Nahpou] =025M Impo = 0.25M ka= -- is very large So, change of water са-гі. (6) Hepay dissorater frotones in water te poy + H₂O + HPO 2 + H₂O + ...üi The expression of eauilibrium. constant (HPO ) [tooth water [H₂ Poyr The reaction (in) showing concern second disociation of broton. So, here, ka should be needed for this reaction. : pka= -log ka her, pkan slog kan --Loq (6.3%988) - 72 & Pkar Pkar = 7.2 Se, o irmat considera to)

() & HPO ? - shou Hapoy & HO . Hot Initially with pure water was present before reaction. (Hitz 0+1 = 10-7m Reaction anotient Q [H3 0+) CH PO?? in Q? - list poul = 107g 0125 023 = 10-77 from left will shift here, kan reauilibrium countans) = 6.3x10-8 So, ay kar So, reaction to right pa - +40 Hz ot & HPO 2 - 0125M - 1017 0.25 the M - ROM (onston) – spotahn E auilibriven contant ko [hothee THPO 2-2 kag is very - ом (0.25 x ) c: & e & auilibrium contan THPO 2 - a ka= The poyder small s 025 th 2025 . 0.25 0.25 050 6.8410 8 10 ore .:2-310-7.469 Xvan 8 ) = 817X20-8 M

in At equilibrium 18+] - Cldba, 0) - 2017 - =fort (3.7x108) 3m 8:3x108m fe) pr-209 [,01] = -. 6? = 7.2