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?

Answer 1

Answer:

1. 0.00352 M

2. 2HNO3(aq) + Sr(OH)2(aq) -----> Sr(NO3)2(aq) + 2H2O(l)

3. 0.00534 M

Explanation:

1.

Mass of strontium hydroxide= 10.45 g

Volume of solution = 41.00 ml

Number of moles = mass of Sr(OH)2/molar mass of Sr(OH)2 = 10.45g/121.63 g/mol= 0.0859 moles

Molarity= number of moles × volume = 0.0859 ×41/1000 = 0.00352 M

2.

2HNO3(aq) + Sr(OH)2(aq) -----> Sr(NO3)2(aq) + 2H2O(l)

3.

Concentration of acid CA= the unknown

Volume of acid VA= 31.5 ml

Concentration of base CB= 0.00352 M

Volume of base VB= 23.9 ml

Number of moles of acid NA= 2

Number of moles of base NB= 1

From;

CAVA/CBVB = NA/NB

CAVANB= CBVBNA

CA= CBVBNA/VANB

CA= 0.00352 × 23.9 ×2/31.5 ×1

CA= 0.00534 M