Mg 1.458 g pH = –0.477 HCl 80.0 mL

Mg 1.458 g pH = –0.477 HCl 80.0 mL

 

 

A 1.458 g of Mg reacts with 80.0 mL of a HCl solution whose pH is –0.477.

The change is pH when all Mg has reacted will be:

 

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Mg의 몰질량 = 24.31 g/mol

1.458 g / (24.31 g/mol) = 0.06 mol Mg

 

 

 

pH = –log[H^+] = –0.477

[H^+] = 10^(0.477) = 3 M = [HCl]

 

 

 

(3 mol/L) (80.0/1000 L) = 0.24 mol HCl

 

 

 

Mg + 2HCl → MgCl2 + H2

 

Mg : HCl = 1 : 2 계수비(= 몰수비) 이므로,

0.06 mol Mg와 반응하는 HCl의 몰수를 계산하면,

Mg : HCl = 1 : 2 = 0.06 mol : ? mol

? = 2 × 0.06 = 0.12 mol HCl

---> HCl은 이만큼 있다, 남는다?

---> HCl은 과잉 반응물, Mg가 한계 반응물.

( 참고: 한계 반응물 https://ywpop.tistory.com/3318 )

 

 

 

반응 후 남아있는 HCl의 몰수

= 0.24 – 0.12 = 0.12 mol

 

 

 

0.12 mol / (80.0/1000 L) = 1.5 M

 

 

 

pH = –log(1.5) = –0.176

 

 

 

pH 변화를 계산하면,

–0.176 – (–0.477) = 0.301

 

 

 

답: 0.3

 

 

 

 

[키워드] Mg sample 1.87 g pH = –0.544 HCl 80.0 mL