Sr(OH)2 12.50 g water 50.00 mL HNO3 37.5 mL 23.9 mL

Sr(OH)2 12.50 g water 50.00 mL HNO3 37.5 mL 23.9 mL

 

 

a) A strontium hydroxide solution is prepared

by dissolving 12.50 g of Sr(OH)2 in water

to make 50.00 mL of solution.

What is the molarity of this solution?

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 37.5 mL aliquot of the nitric acid solution,

what is the concentration (molarity) of the acid?

 

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a)

Sr(OH)2의 몰질량 = 121.63 g/mol 이므로,

12.50 g / (121.63 g/mol) = 0.10277 mol Sr(OH)2

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

몰농도 = 용질 mol수 / 용액 L수

( 참고 https://ywpop.tistory.com/3222 )

 

= 0.10277 mol / (50.00/1000 L)

= 2.0554 M

≒ 2.06 M

 

 

 

b)

2HNO3(aq) + Sr(OH)2(aq) → Sr(NO3)2(aq) + 2H2O(l)

2HNO3 + Sr(OH)2 → Sr(NO3)2 + 2H2O

 

 

 

c)

aMV = bM’V’

( 참고 https://ywpop.tistory.com/4689 )

 

(1) (? M) (37.5 mL) = (2) (2.06 M) (23.9 mL)

 

? = [(2) (2.06) (23.9)] / [(1) (37.5)] = 2.63 M

 

 

 

또는

(2.06 M) (23.9 mL) = 49.234 mmol Sr(OH)2

 

HNO3 : Sr(OH)2 = 2 : 1 = ? mmol : 49.234 mmol

 

? = 2 × 49.234 = 98.468 mmol HNO3

 

98.468 mmol / 37.5 mL = 2.63 M HNO3