PH Changes in Buffers Part A What is the pH of a bufer pepared by adding 0 508 m
ID: 1025401 • Letter: P
Question
PH Changes in Buffers Part A What is the pH of a bufer pepared by adding 0 508 mol of the weak ad HA to 0.608 mol of NaA 2.00 of solution? The dasociation constiant Ko HA's M6 × 10', When a soluion contains a weak acid and its conjugate base or weak base nd its coryugate acid, be a buner soluton ufes resist change in pH following the addtion of acid or base A buffer solution prepared from a wesk acid (ELA) and ts conjugate base (A )is represented as Express the pH numerically to thrse decimal places HA(g)H' (4) +A (ag) The btw wn foilow LeChataler's prople "acid "dded, te macion shifts to consume the added I' forming more HA when base is added, base wil act with H"duang its dissociation of HA into H tends to remain and A In both instances H The pH of a buter is cnluialed by using the Henderson Hasselbaich equation Part B pH=pK, + log What is the pH aher 0 150 mol of HC s added to the buffler from Part A? Assume no volume change on the addition of the ackd Express the pt numerically to three declimal places View Available Hin pH. Part C What is the pll aor 0195 mol of NaO is added to the bur om Prt A? Assume no voume change on the adsition of the bas Express the pH numerically to thrse decimal places View Availab. Hits)Explanation / Answer
pH of Buffer solution
Part A :-Given ; Ka = 5.66 (10-7); HA (weak acid)= 0.506 mol ; NaA (conjugate base)= 0.606 mol ; volume = 2.0 L
pKa = -log Ka = -log 5.66 (10-7)
= -[0.752+(-7)]
= -0.752 - (-7)
= -0.752 + 7
pKa = 6.215
Formula :- pH = pKa + log [A-] / [HA]
= 6.215 + log [0.606] / [0.506]
= 6.215 + log 1.197
= 6.215 + 0.078
pH = 6.293
Part B :- 0.150 mol HCl is added to part A
HA = 0.506 + 0.150 = 0.656
NaA = 0.606 - 0.150 = 0.456
pH = pKa + log [A-] / [HA]
= 6.215 + log [0.456] / [0.656]
= 6.215 + log 0.695
= 6.215 +(- 0.158)
pH = 6.057
Adding the HCl an acid has changed the pH of the buffer in the acidic direction, from 6.293 to 6.057.
Part C :- 0.195 mol NaOH is added to part A
HA = 0.506 - 0.195 = 0.311
NaA = 0.606 + 0.195 = 0.801
pH = pKa + log [A-] / [HA]
= 6.215 + log [ 0.801] / [0.311]
= 6.215 + log 2.57
= 6.215 + 0.410
pH = 6.625
Adding the NaOH base has changed the pH of the buffer in the basic direction, from 6.293 to 6.625.
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