0.5413 g 1.396 M HCl 10.00 mL CO2 0.1004 M NaOH 39.96 mL mass percent

0.5413 g 1.396 M HCl 10.00 mL CO2 0.1004 M NaOH 39.96 mL mass percent

 

 

석회석에 들어 있는 방해석의 무게 백분율을 구하시오.

 

 

Limestone consists mainly of the mineral calcite, CaCO3.

The carbonate content of 0.5413 g of powdered limestone

was measured by suspending the powder in water,

adding 10.00 mL of 1.396 M HCl,

and heating to dissolve the solid and expel CO2:

CaCO3(100.087 g/mol) + 2H^+ → Ca^2+ + CO2 + H2O

 

The excess acid required 39.96 mL of 0.1004 M NaOH

for complete titration to a phenolphthalein end point.

Find the weight percent of calcite in the limestone.

 

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▶ 참고: 역정적법

[ https://ywpop.tistory.com/19258 ]

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과량으로 가한 HCl의 몰수를 계산하면,

(1.396 mol/L) (10.00/1000 L) = 0.01396 mol H^+

---> 이것을 NaOH와 CaCO3가 나눠서 반응

 

 

 

(0.1004 mol/L) (39.96/1000 L) = 0.004012 mol OH^-

---> NaOH가 반응한 양

 

 

 

0.01396 – 0.004012 = 0.009948 mol H^+

---> CaCO3가 반응한 양

 

 

 

CaCO3 + 2H^+ → Ca^2+ + CO2 + H2O

 

CaCO3 : H^+ = 1 : 2 = ? mol : 0.009948 mol

 

? = 0.009948 / 2 = 0.004974 mol CaCO3

 

 

 

0.004974 mol × (100.087 g/mol) = 0.4978 g CaCO3

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

(0.4978 g / 0.5413 g) × 100 = 91.96%

---> weight percent of calcite in the limestone

 

 

 

 

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