CrO4^2- 0.250 L CaCrO4 314.1 mg Ca^2+

CrO4^2- 0.250 L CaCrO4 314.1 mg Ca^2+

Ca^2+(aq) + CrO4^2-(aq) → CaCrO4(s)

 

 

농도를 모르는 칼슘 용액 0.250 L를

과량의 크로뮴산 암모늄과 결합시켰다.

이때 생긴 크로뮴산 칼슘 침전의 질량은 314.1 mg이었다.

Ca^2+의 몰농도는?

 

 

Concentration of calcium in a sample may be determined

by precipitation using the chromate ion, CrO4^2-.

Ca^2+(aq) + CrO4^2-(aq) → CaCrO4(s)

A chemist combined 0.250 L of an unknown calcium solution

with an excess of ammounium chromate.

This resulted in the precipitation of calcium chromate.

The mass of the precipitate was 314.1 mg.

What was the molar concentration of Ca^2+ in the original sample?

 

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CaCrO4의 몰질량 = 156.07 g/mol 이므로,

(314.1 mg) (1 g / 1000 mg) (1 mol / 156.07 g)

= (314.1) (1 / 1000) (1 / 156.07) = 0.002013 mol CaCrO4

 

 

 

Ca^2+(aq) + CrO4^2-(aq) → CaCrO4(s)

 

Ca^2+ : CaCrO4 = 1 : 1 계수비(= 몰수비) 이므로,

Ca^2+의 몰수 = 0.002013 mol

 

 

 

[Ca^2+] = 0.002013 mol / 0.250 L

= 0.008052 mol/L

= 0.008052 M

 

 

 

답: 8.052×10^(-3) M

 

 

 

 

[공식] Ca^2+(aq)의 몰농도(M)

= CaCrO4의 질량(mg) / 1000 / 156.07 / 시료의 부피(L)

= 314.1 / 1000 / 156.07 / 0.250

= 0.008050 M