Question

a) A strontium hydroxide solution is prepared by dissolving 10.55 g of Sr(OH)2 in water to make 59.00 mL of solution.What is the molarity of this solution? (1.470M is the right answer)

b) Next the strontium hydroxide solution prepared in part (a) is used to titrate a nitric acid solution of unknown concentration.Write a balanced chemical equation to represent the reaction between strontium hydroxide and nitric acid solutions.

c) If 23.9 mL of the strontium hydroxide solution was needed to neutralize a 31.5 mL aliquot of the nitric acid solution, what is the concentration (molarity) of the acid?

Could you please include steps to parts a-c? Thank you!

Answer 1

a) Molarity=(weight/molecular weight)(1/V(L))

Weight of Sr(OH)2=10.55 g, Volume=59 mL=0.059 L, molecular weight of Sr(OH)2=121.6347 g/mol

Molarity=(10.55g/121.6347 g/mol)(1/0.059 L)=1.47 mol/L=1.47 M.

Molarity of Sr(OH)2=1.47 M.

b) The balanced chemical equation is

Sr(OH)2 + 2 HNO3 → Sr(NO3)2 + 2 H2O

c) In nuetralization reaction, mol of acid=mol of base

Moles of Sr(OH)2 in 23.9 mL=(23.9*10-3 L*1.47 mol/L)(2 mol OH^-/1mol Sr(OH)2)=0.07026 mol OH-,

At nuetralization,mol OH-=mol H+, 0.07026 mol OH-

Conc.HNO3=0.07026 mol/0.0315 L=2.23 M

Conc. HNO3=2.23M.