산-염기 중화반응. HCl + Mg(OH)2

산-염기 중화반응. HCl + Mg(OH)2

 

 

0.102 g Mg(OH)2와 0.0600 M HCl 30.0 mL를 혼합한 용액의 pH

 

What is the pH of the solution that results when 0.102 g of Mg(OH)2 is mixed with 30.0 mL of 0.0600 M HCl?

 

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Mg(OH)2의 몰질량 = 58.32 g/mol 이므로,

0.102 g / (58.32 g/mol) = 0.00175 mol Mg(OH)2

 

Mg(OH)2(aq) → Mg^2+(aq) + 2OH^-(aq)

Mg(OH)2 1몰당 OH^- 2몰 생성되므로, OH^-의 몰수

= 2 * 0.00175 mol = 0.00350 mol OH^-

 

0.0600 M HCl 30.0 mL에 들어있는 HCl(=H^+)의 몰수

= (0.0600 mol/L) * (30.0/1000 L) = 0.00180 mol H^+

 

OH^- > H^+ 이므로, 중화반응 후 남아있는 OH^-의 몰수

= 0.00350 – 0.00180 = 0.00170 mol OH^-

 

[OH^-] = 0.00170 mol / (30.0/1000 L) = 0.0567 M OH^-

 

pH = 14 – pOH = 14 - (-log(0.0567)) = 12.8

 

 

답: pH = 12.8

 

 

[ 관련 예제 http://ywpop.tistory.com/3693 ]

 

 

 

What is the pH of the solution that results when 0.102 g of Mg(OH)2 is mixed with 75.0 mL of 0.0600 M HCl?
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0.102 g / (58.32 g/mol) = 0.00175 mol Mg(OH)2
0.00175 mol * 2 = 0.0035 mol OH^-

 

(0.0600 mol/L) * (75.0/1000 L) = 0.0045 mol H^+

 

0.0045 - 0.0035 = 0.001 mol H^+
(0.001 mol) / (75.0/1000 L) = 0.0133 M H^+

 

pH = -log(0.0133) = 1.88