| ⇦ | 
| ⇨ |
What is the [OH⁻] in the final solution prepared by mixing 20 mL of 0.05 M Hcl with 30 mL of 0.1 M Ba(OH)₂?
Options
(a) 0.4 M
(b) 0.005 M
(c) 0.12 M
(d) 0.1 M
Correct Answer:
0.1 M
Explanation:
No of milliequivalent of HCl = 20 x 0.05 =1.0
No of milliequivalent of Br(OH)₂ = 30 x 0.1 x 2 = 60
after neutralization, no of milliequivalents in 50 ml of solution = 6-1=5
Total vol of solution = 20 + 30 = 50 ml
No of milliequivalent of OH⁻ is 5 in 50 ml
[OH⁻] = (5 x 100 / 10 ) x 10⁻³ = 0.1 M
Related Questions: - The lattice energy order for lithium halide is
- The products of the following reaction are
- Chlorine is liberated, when we heat
- The compound that gives both iodoform and Fehling’s tests is
- At when pressure will a quantity of gas, which occupies 100 mL at a pressure of 720 mm
Question Type: Memory
(964)
Difficulty Level: Easy
(1008)
Topics: Basic Concepts of Chemistry
(94)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The lattice energy order for lithium halide is
- The products of the following reaction are
- Chlorine is liberated, when we heat
- The compound that gives both iodoform and Fehling’s tests is
- At when pressure will a quantity of gas, which occupies 100 mL at a pressure of 720 mm
Question Type: Memory (964)
Difficulty Level: Easy (1008)
Topics: Basic Concepts of Chemistry (94)
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
Leave a Reply