weak acid 100.0 mL 0.09381 M NaOH 27.63 mL pH 10.99 16.47 mL

weak acid 100.0 mL 0.09381 M NaOH 27.63 mL pH 10.99 16.47 mL

약산 100.0 mL를 0.09381 M NaOH로 적정할 때,

당량점에 도달하는데 27.63 mL가 소모되었으며

당량점에서의 pH는 10.99였다.

NaOH 용액 16.47 mL만을 첨가시켰을 때,

pH는 얼마인지 계산하시오.

When 100.0 mL of a weak acid were titrated with 0.09381 M NaOH, 27.63 mL were required to reach the equivalence point. The pH at the equivalence point was 10.99.

What was the pH when only 16.47 mL of NaOH had been added?

[참고] 약산-강염기 적정

[ https://ywpop.tistory.com/2736 ]

[ https://ywpop.tistory.com/15933 ]

MV = M’V’

M(산 농도) × V(산 부피) = M’(염기 농도) × V’(염기 부피)

( 참고 https://ywpop.tistory.com/4689 )

M = M’V’ / V

= (0.09381 M) (27.63 mL) / (100.0 mL)

= 0.02592 M

---> HA의 몰농도

당량점에서 전체 용액의 부피

= 100.0 + 27.63

= 127.63 mL

당량점에서 짝염기(A^-)의 몰농도

= (0.02592 mol/L) (100.0/1000 L) / (127.63/1000 L)

= (0.02592) (100.0/1000) / (127.63/1000)

= 0.02031 mol/L

= 0.02031 M = [A^-]

당량점에서 짝염기의 가수분해

A^-(aq) + H2O(l) ⇌ HA(aq) + OH^-(aq)

Kh = Kb = [HA] [OH^-] / [A^-]

= (x) (x) / (C–x)

pOH = 14 – pH

= 14 – 10.99

= 3.01

[OH^-] = 10^(-3.01)

= 9.77×10^(-4) M

C–x ≒ C 라 근사처리하면,

Kb = x^2 / C

= (9.77×10^(-4))^2 / 0.02031

= 4.70×10^(-5)

Ka = Kw / Kb

= (1.0×10^(-14)) / (4.70×10^(-5))

= 2.13×10^(-10)

pKa = –logKa

= –log(2.13×10^(-10))

= 9.67

What was the pH when only 16.47 mL of NaOH had been added?

(0.02592 mol/L) (100.0/1000 L)

= 0.002592 mol HA

(0.09381 mol/L) (16.47/1000 L)

= 0.001545 mol NaOH

HA(aq) + OH^-(aq) → A^-(aq) + H2O(l)

약산에 강염기를 가하면,

가한 강염기의 몰수만큼,

약산의 몰수는 감소(소모),

짝염기의 몰수는 증가(생성).

( 참고 https://ywpop.tistory.com/15932 )

0.002592 – 0.001545

= 0.001047 mol HA

Henderson-Hasselbalch 식

pH = pKa + log([짝염기]/[약산])

( 참고 https://ywpop.tistory.com/1926 )

= 9.67 + log(0.001545 / 0.001047)

= 9.84

What was the pH when only 19.47 mL of NaOH had been added?

(0.02592 mol/L) (100.0/1000 L)

= 0.002592 mol HA

(0.09381 mol/L) (19.47/1000 L)

= 0.001826 mol NaOH

0.002592 – 0.001826

= 0.000766 mol HA

pH = pKa + log([짝염기]/[약산])

= 9.67 + log(0.001826 / 0.000766)

= 10.05

[참고] 당량점에서 짝염기(A^-)의 몰농도 계산 설명

처음(초기) 산의 몰수만큼 짝염기가 생성되지만,

(0.02592 mol/L) (100.0/1000 L)

= 0.002592 mol

적정 과정에서 NaOH 용액이 가해졌기 때문에,

당량점에서 용액의 부피는 처음 산 용액의 부피와 다르다.

약산 100.0 mL + NaOH 27.63 mL

= 127.63 mL

---> 당량점에서 용액의 부피

이 때문에,

0.002592 mol / (127.63/1000 L)

= 0.02031 M

한 번에 계산하면,

(0.02592 mol/L) (100.0/1000 L) / (127.63/1000 L)

= 0.02031 M

[키워드] 약산-강염기 적정 기준, When 100.0 mL of weak acid was titrated with 0.09381 M NaOH, 27.63 mL was consumed to reach the equivalence point, and the pH at the equivalence point was 10.99. Calculate the pH when only 16.47 mL of NaOH solution is added.