CHM 1032 Exam 3 Name Show all work, cancel units and use significant figures. Th

Question

CHM 1032 Exam 3 Name Show all work, cancel units and use significant figures. The titration of a 75.00 mL sample of sulfurous acid solution of unknown concentration requires 150.00 of a 0.175 M lithium hydroxide solution to reach the equivalence point. Write out the balance equation for this acid/base titration including all phases. What is the molarity of the acid in this titration? At the end of the titration what is the molarity of the salt? Assume the acid and base solutions volumes are additive

Explanation / Answer

Q1

a)

balanced reaction between

H2SO3(aq) + LiOH(aq) = Li2SO3(aq) + H2O(l)

balance

H2SO3(aq) + 2LiOH(aq) = Li2SO3(aq) + 2H2O(l)

b)

find molarity of acid

mmol of base

mmol of base = Mbase*Vbase = 0.175 * 150 = 26.25 mmol of base

now..

mmol fo acid = 1/2*mmol of base = 26.25/2 = 13.125 mmol

[H2SO3] = mmol/V = 13.125/75 = 0.175 M

c)

Molarity of salt is

Li2SO3 --> 13.125 mmol forms

Vtotal = 75+150 = 225 mL

[Li2SO3] = mmol/V = 13.125 / (225) = 0.05833 M

d)

find m/v % of salt

mass of Li2SO3 = mmol/10^3 * MW = (13.125 /1000)(93.9452 ) = 1.233 g

Vtotal = 225 mL

%m/v = mass / volume * 100% = 1.233/225*100 = 0.548%

e)

water molecules created

mmol of H2O = 2*LiSO4 = 13.125 *2 = 26.5 mmol = 26.5*10^-3 mol

1 mol = 6.022*10^23 moelcules

26.5*10^-3 mol= (26.5*10^-3 )*6.022*10^23 moelcules = 1.59*10^22 molecules

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