A strontium hydroxide solution is prepared by dissolving 10.40 g of Sr(OH)2 in w

Question

A strontium hydroxide solution is prepared by dissolving 10.40 g of Sr(OH)2 in water to make 47.00 mL of solution.What is the molarity of this solution?

Express your answer to four significant figures and include the appropriate units.

SubmitMy AnswersGive Up

Part 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.

Express your answer as a chemical equation. Identify all of the phases in your answer.

Part 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?

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

Explanation / Answer

m = 10.4 g of Sr(OH)2

MW = 121.63

mol = mass/MW = 10.4/121.63 = 0.08550 mol of Sr(OH)2

V = 47 ml = 47/1000 L = 0.047 L

then

[Sr(OH)2] = mol/L = 0.08550 /0.047 = 1.81914 M

b)

HNO3 + Sr(OH)2 = H2O + Sr(NO3)2

the balanced equation

2HNO3 + Sr(OH)2 = 2H2O + Sr(NO3)2

c)

V = 23.9 ml of Sr(OH)2

V = 31.5 mL

then [HNO#] = ?

mol of Bas e= MV = 23.9*1.81914 = 43.477446 mmol of Sr(OH)2

hen we require twice of acid

43.477446*2 = 86.954892 mmol of acid

[Acic] = mmol/ml = 86.954892 / (31.5) = 2.76047 M

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.