Part e:Calcualate the pH at the equivalence point of the titration(50.00mL of ac
ID: 527366 • Letter: P
Question
Part e:Calcualate the pH at the equivalence point of the titration(50.00mL of acid and 37.21 mL of base). Show the neutralization reaction, the equilibrium expression for the weak base hydrolysis, the mass action express, state any assumptions used in your calculations, and show all of your work.
Please answer a-e
A student titrated 50.00 mL of an unknown monoprotic acid, "HA" with standard 02792 M NaOH. Use the experimental results given below to calculate the following solution parameters. (a (5pts If 3721 ml, of the standard base was required to reachthetitration endpoint, oaleulate dhe Initial of the unknown acid. Show your work and include a balanced equation for the zRdlivreaction. Ma 0,2018 M 0.03 (b) (5pts) A student next pipetted 25.00 mL of the unknown acid into a beaker and with a z 9.30 the standard base. The resulting solution 5320 as measured unknown acid. all your work including the o meter. the form of the mass action expression. weak acid equilibrium reaction and PH 5.320 s 530Explanation / Answer
(a) moles of base used = 0.2792 M x 37.21 ml = 10.39 mmol
initial molarity of acid solution = 10.39 mmol/50 ml = 0.2078 M
(b) moles base added = 0.2792 M x 9.30 ml = 2.5966 mmol
moles acid neutralized = 2.5966 mml
moles acid remained = 0.2078 M x 25 ml - 2.5966 = 2.5984 mmol
pH = pKa + log(base/acid)
5.320 = pKa + log(2.5966/2.5984)
pKa = 5.3203
pKa = -log[Ka]
Ka = 4.783 x 10^-6
(c) Initial molarity of acid (HA) = 0.2078 M
HA + H2O <==> A- + H3O+
let x amount has dissociated
Ka = [A-][H3O+]/[HA]
4.783 x 10^-6 = x^2/0.2078
x = [H3O+] = 9.97 x 10^-4 M
pH = -log[H3O+] = 3.00
(d) Kb = Kw/Ka
= 1 x 10^-14/4.783 x 10^-6
= 2.091 x 10^-9
(e) HA + B <==> BH+ + A-
molarity of A- formed at equivalence point = 0.2078 M x 50 ml/(50 + 37.21) ml = 0.12 M
A- + H2O <==> HA + OH-
let x amount has hydrolyzed
Kb = [HA][OH-]/[A-]
2.091 x 10^-9 = x^2/0.12
x = [OH-] = 1.58 x 10^-5 M
pOH = -log[OH-] = 4.80
pH = 14 - pOH = 9.20