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 oxidation number of sulphur is -1 in
2. Which of the following is the IUPAC name of the compound
3. Identify the correct order of the size of the following elements:
4. Phenyl isocyanide test is used to identify
5. The standard electrode potential is measured by

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