Question

1a, A strontium hydroxide solution is prepared by dissolving 10.45 g of Sr(OH)2 in water to make 41.00 mL of solution.What is the molarity of this solution?

1b. 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.

1c. 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? Express your answer to three significant figures and include the appropriate units.

Answer 1

1A)

Molar Mass of Sr(OH) ₂ = 87.62 + ( 2 x 16.00) + ( 2 x 1.0079) = 121.64 g / mol

We have, No of moles = Mass / Molar mass

= 10.45 g / ( 121.64 g / mol )

= 0.08591 mol

We know that , Molarity = No of moles of solute / Volume of solution in L

Therefore, Molarity of Sr(OH) ₂ = 0.08591 mol / 0.041 L

= 2.10 M

1b) Sr(OH) _{2 (aq)}+ 2 HNO_3(aq) Sr(NO₃)_{2 (aq)}+ 2 H₂O _(l)

1c) From above reaction , Stoichiometric ratio = no of moles of acid / no of moles of Base= 2 /1=2

We have relation ,

M _acid x V _acid = M _base x V _base x Stoichiometric ratio

M _acid = M _base x V _base x Stoichiometric ratio / V _acid

= 2.10 M x 23.9 ml x 2 / 31.5 ml

= 3.19 M

ANSWER : Molarity of Nitric acid = 3.19 M