A 2.000 grams of a salt are dissolved in enough water to make a solution with a

Question

A 2.000 grams of a salt are dissolved in enough water to make a solution with a volume of 250.0 mL. The solution is titrated with a 0.1111 M sodium hydroxide solution and reached the equivalence point at 30.00 mL. The pH of the analyze solution was determined to be 9.1989 after the addition of 10.00 ml of the sodium hydroxide and 9.8010 after the addition of 20.00 ml of the sodium hydroxide. a) what is the molecular weight of the salt assuming it is monoprotic b) is the salt acidic or basic? c) what is the K of the acid form of the salt d) what was the pH before addition of base?

Explanation / Answer

a)

m = 2 g of salt

V = 250 mL of salt

NaOH --> 0.1111 M of NaOH, V = 30 mL

pH of anylite determined after 10mL of base --> 9.1989 and 20 mL --> 9.8010

a)

find molar mass if monoprotic

so

mmol of base = MV = 0.1111*30 = 3.333 mmol of base

so

mmol of acid = 3.333 mmol = 3.333*10^-3 mol

MW = mass/mol = 2/(3.333*10^-3) = 600.06 g/mol

b)

this salt is clearly ACIDIC, since it is reacting with a STRONG BASE

c)

for Ka, apply buffer equation

mmol of acid initially = 3.333

mmol of base added ( 10mL) = 0.1111*10 = 1.111

after neutralization

mmol of acid left = 3.333 -1.111 = 2.222

mmol of conjguate formed = 0 + 1.111 = 1.111

apply buffer equation

pH = pKa + log(conjugate/acid)

9.1989 = pKa + log(1.111/2.222)

pKa = 9.1989 - log(1.111/2.222) = 9.49992 = 9.50 approx

Ka = 10^-pKa = 10^-9.5 =3.16*10^-10

d)

pH before any addition

HA <->H+ + A-

Ka = [H+][A-]/[HA]

3.16*10^-10 = x*x/(3.333/250 - x)

x = 2.05*10^-6

pH = -log(x) = -log( 2.05*10^-6) = 5.69, which is acidic, as expected

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