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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: - Alkali metals in each period have
- The IUPAC name of (CH₃)₂CH – CH₂ – CH₂Br is
- For the reaction R → P a graph of [R] against time is found to be a straight
- Which of the following is the monoliquid propellant
- The spin only magnetic moment of [CrF₆]⁴⁻ (atomic number for Cr is 24) is
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
- Alkali metals in each period have
- The IUPAC name of (CH₃)₂CH – CH₂ – CH₂Br is
- For the reaction R → P a graph of [R] against time is found to be a straight
- Which of the following is the monoliquid propellant
- The spin only magnetic moment of [CrF₆]⁴⁻ (atomic number for Cr is 24) is
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
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