0.0800 M HCl 20.00 mL MCO3 0.1022 g 0.1000 M NaOH 5.64 mL

0.0800 M HCl 20.00 mL MCO3 0.1022 g 0.1000 M NaOH 5.64 mL

 

 

The molar mass of a certain metal carbonate, MCO3,

can be determined by adding an excess of HCl acid

to react with all the carbonate and then

“back-titrating” the remaining acid with NaOH.

(a) Write an equation for these reactions.

(b) In a certain experiment, 20.00 mL of 0.0800 M HCl

was added to a 0.1022 g sample of MCO3.

The excess HCl required 5.64 mL of 0.1000 M NaOH for neutralization.

Calculate the molar mass of the carbonate and identify M.

 

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(0.0800 mol/L) (20.00/1000 L) = 0.0016 mol HCl

= MCO3와 반응한 HCl의 몰수 + NaOH와 반응한 HCl의 몰수

---> 이것이 “역적정”의 핵심.

 

 

 

HCl + NaOH → NaCl + H2O

 

(0.1000) (5.64/1000) = 0.000564 mol

= NaOH와 반응한 HCl의 몰수

 

 

 

2HCl + MCO3 → MCl2 + H2CO3

 

0.0016 – 0.000564 = 0.001036 mol

= MCO3와 반응한 HCl의 몰수

 

 

 

HCl : MCO3 = 2 : 1 몰수비로 반응하므로,

( 참고 https://ywpop.tistory.com/4711 )

 

HCl : MCO3 = 2 : 1 = 0.001036 mol : ? mol

 

? = 0.001036 / 2 = 0.000518 mol MCO3

---> 0.001036 mol HCl과 반응한 MCO3의 몰수

 

 

 

MCO3의 몰질량을 계산하면,

0.1022 g / 0.000518 mol = 197.30 g/mol

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

CO3의 몰질량 = 60.01 g/mol 이므로,

197.30 – 60.01 = 137.29 g/mol

---> M의 몰질량

 

 

 

주기율표에서 137.29 g/mol인 원소를 찾으면,

Ba(바륨).

---> MCO3 = BaCO3