(NH4)2SO4 10.8 g water 100 mL 10 mL 50 mL

(NH4)2SO4 10.8 g water 100 mL 10 mL 50 mL

 

 

황산암모늄 10.8 g을 물에 녹여 100 mL 저장 용액을 만들었다.

 

A solution is prepared by dissolving 10.8 g ammonium sulphate

in enough water to make 100 mL of stock solution.

A 10 mL sample of this stock solution is added to 50 mL of water.

Calculate concentration of ammonium ions and sulphate ions in the final solution?

 

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(NH4)2SO4의 몰질량 = 132.14 g/mol 이므로,

10.8 g / (132.14 g/mol) = 0.08173 mol (NH4)2SO4

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

몰농도 = 0.0817 mol / 0.1 L

= 0.817 mol/L

---> 100 mL stock solution의 몰농도

 

 

 

(0.817 mol/L) (10/1000 L) = 0.00817 mol (NH4)2SO4

---> 10 mL stock solution에 들어있는 (NH4)2SO4의 몰수

 

 

 

0.00817 mol / (60/1000 L) = 0.136 mol/L (NH4)2SO4

---> 묽힌 stock solution의 몰농도

 

 

 

(NH4)2SO4(aq) → 2NH4^+(aq) + SO4^2-(aq)

(NH4)2SO4 : NH4^+ : SO4^2- = 1 : 2 : 1 계수비(= 몰수비)

 

답:

[NH4^+] = 2 × 0.136 = 0.272 M

[SO4^2-] = 0.136 M