When 10 mL of 0.1 M acetic acid (pKa = 5.0 ) is titrated against 10mL of 0.1 M

⇦

⇨

When 10 mL of 0.1 M acetic acid (pKa = 5.0 ) is titrated against 10mL of 0.1 M ammonia solution (pKb = 5.0),the equivalence point occurs at pH

Options

(a) 5
(b) 6
(c) 7
(d) 9

Correct Answer:

7

Explanation:

pKₐ = -logKₐ : pK(b) = -logK(b),
pH = -1/2[logKₐ + log K(w) – logK(b),
-1/2[-5 + log(1*10⁻¹⁴)-(-5)],
-1/2[-5-14+5]=-1/2(-14)=7.

Related Questions:

1. The correct order of acidic strength is
2. Which of the following isomeric alcohols react fastest with HCL
3. The pressure and temperature of 4 dm³ of carbon dioxide gas are doubled
4. The ion which is not tetrahedral in shape is
5. Grignard reagent is prepared by the reaction between

Topics: Equilibrium (104)
Subject: Chemistry (2512)

---

Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score

18000+ students are using NEETLab to improve their score. What about you?

Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.

---

Share this page with your friends