0.2 M HNO3 용액의 pH. 0.002 M H2SO4 용액의 pH

0.2 M HNO3 용액의 pH. 0.002 M H2SO4 용액의 pH

 

 

What is the pH of a 0.2 mol/dm3 nitric acid and a 0.002 mol/dm3 of sulfuric acid (assume as it is a strong, diprotic acid) solution?

 

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HNO3(aq) → H^+(aq) + NO3^-(aq)

[HNO3] = [H^+] = 0.2 M

 

pH = -log[H^+] = -log(0.2) = 0.699

 

 

H2SO4(aq) → 2H^+(aq) + SO4^2-(aq)

H2SO4 : H^+ = 1 : 2 계수비(= 몰수비) 이므로,

(assume as it is a strong, diprotic acid)

[H^+] = 2 × 0.002 M = 0.004 M

 

pH = -log[H^+] = -log(0.004) = 2.398

 

 

답: pH = 0.699, pH = 2.398

 

 

[참고] 1 mol/dm3 = 1 mol/L = 1 M
1 dm3 = 1 L
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