What is the [OH$$-$$] in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of 0.10 M Ba(OH)2?
A
0.40 M
B
0.0050 M
C
0.12 M
D
0.10 M
Explanation
Number of equivalents of
H+ = 20.0 × 0.050 milliequivalents
= 1.0 milliequivalents
Number of equivalents of OH–
= 2 × 30.0 × 0.10
= 6.0 milliequivalents
$$ \therefore $$ Equivalents of OH-
left after neutralization
= 6 – 1 = 5 milliequivalents
Total volume after neutralization
= 20.0 + 30.0 mL
= 50 mL
[OH-] = $${5 \over {50}}$$ = 0.1 M
4
AIPMT 2009
MCQ (Single Correct Answer)
The dissociation constants for acetic acid and HCN at 25oC are 1.5 $$ \times $$ 10$$-$$5 and 4.5 $$ \times $$ 10$$-$$10 respectively. The equilibrium constant for the equilibrium
CN$$-$$ + CH3COOH $$\rightleftharpoons$$ HCN + CH3COO$$-$$ would be