3.00 g salicylic acid in 50.0 mL of 0.1130 M NaOH

3.00 g salicylic acid in 50.0 mL of 0.1130 M NaOH

 

 

Calculate the pH of a solution that is prepared by dissolving 3.00 g salicylic acid, (C6H4(OH)COOH (138.12 g/mol, Ka = 1.06×10^-3) in 50.0 mL of 0.1130 M NaOH and diluting to 500.0 mL. (The Henderson-Hasselbalch equation will not work in this case).

 

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Let, salicylic acid = HA

 

HA의 mol수 = 3.00 g / (138.12 g/mol) = 0.02172 mol

( 계산 설명 http://ywpop.tistory.com/7738 )

 

NaOH의 mol수 = (0.1130 mol/L) × 0.0500 L = 0.00565 mol

 

HA(aq) + NaOH(aq) → NaA(aq) + H2O(l)

 

HA의 mol수 > NaOH의 mol수 이므로,

NaOH의 mol수만큼 HA는 중화(소모)되고, NaA는 생성된다.

 

0.02172 – 0.00565 = 0.01607 mol HA

0.01607 mol / 0.5000 L = 0.03214 M HA

 

0.00565 mol / 0.5000 L = 0.01130 M NaA

 

 

HA(aq) ⇌ H^+(aq) + A^-(aq)

Ka = [H^+][A^-] / [HA]

 

[H^+] = Ka[HA] / [A^-]

= (1.06×10^(-3))(0.03214) / (0.01130)

= 0.003015 M

 

pH = -log(0.003015) = 2.521

 

즉, Henderson-Hasselbalch equation으로 계산하면,

pH = pKa + log([A^-]/[HA])

= -log(1.06×10^(-3)) + log(0.01130/0.03214) = 2.521

 

 

However, [H^+] is not << [HA] and [NaA]

[HA] = 0.03214 – [H^+] + [OH^-]

[A^-] = 0.01130 + [H^+] - [OH^-]

 

at this pH, [OH^-] will be negligible.

 

Ka = [H^+][A^-] / [HA]

1.06×10^-3 = [H^+](0.01130 + [H^+]) / (0.03214 – [H^+])

 

0.01130[H^+] + [H^+]^2 = (3.407×10^-5) - (1.06×10^-3)[H^+]

[H^+]^2 + 0.01236[H^+] - (3.407×10^-5) = 0

 

근의 공식으로 [H^+]를 계산하면,

[H^+] = 2.321×10^-3

pH = -log(2.321×10^-3) = 2.634

 

 

답: pH = 2.63