1
MCQ (Single Correct Answer)

AIPMT 2014

Using the Gibb's energy change, $$\Delta $$Go = +63.3 kJ, for the following reaction,

Ag2CO3(s) $$\rightleftharpoons$$ 2 Ag+(aq) + CO32$$-$$ (aq)
the Ksp of Ag2CO3(s) in water at 25oC is
(R = 8.314 J K$$-$$1 mol$$-$$1)
A
3.2 $$ \times $$ 10$$-$$26
B
8.0 $$ \times $$ 10$$-$$12
C
2.9 $$ \times $$ 10$$-$$3
D
7.9 $$ \times $$ 10$$-$$2

Explanation

We know, $$\Delta $$Go = – 2.303 RT log Ksp

$$ \therefore $$ 63300 = – 2.303 × 8.314 × 298 log Ksp

$$ \Rightarrow $$ log Ksp = -11.09

$$ \Rightarrow $$ Ksp = 10-11.09

= 8.0 × 10–12
2
MCQ (Single Correct Answer)

AIPMT 2014

Which of the following salts will give highest pH in water?
A
KCl
B
NaCl
C
Na2CO3
D
CuSO4

Explanation

Na2CO3 is salt of strong base, NaOH and weak acid, H2CO3 hence the pH value of the solution will be high.
3
MCQ (Single Correct Answer)

NEET 2013 (Karnataka)

The dissociation constant of weak acid is 1 $$ \times $$ 10$$-$$4. In order to prepare a buffer solution with a pH = 5, the [Salt]/[Acid] ratio should be
A
4 : 5
B
10 : 1
C
5 : 4
D
1 : 10

Explanation

Given, Ka = 1 × 10–4

pKa = – log (1× 10–4) = 4

pH = pKa + log $${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$

$$ \Rightarrow $$ 5 = 4 + log $${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$

$$ \Rightarrow $$ log $${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$ = 1

$$ \Rightarrow $$ $${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$ = 10 = 10 : 1
4
MCQ (Single Correct Answer)

NEET 2013 (Karnataka)

The values of Ksp of CaCO3 and CaC2O4 are 4.7 $$ \times $$ 10$$-$$9 and 1.3 $$ \times $$ $$-$$9 respectively at 25oC. If the mixture of these two is washed with water, what is the concentration of Ca2+ ions in water ?
A
5.831 $$ \times $$ 10$$-$$5 M
B
6.856 $$ \times $$ 10$$-$$5 M
C
3.606 $$ \times $$ 10$$-$$5 M
D
7.746 $$ \times $$ 10$$-$$5 M

Explanation

CaCO3 $$ \to $$ Ca2+ + CO32-
x x

CaC2O4 $$ \to $$ Ca2+ + C2O42-
y y


[Ca2+] = x + y

Now, Ksp (CaCO3) = [Ca2+] [CO3 2-]

or 4.7 × 10–9 = (x + y)x .......(1)

similarly, Ksp (CaC2O4) = [Ca2+] [C2O42–]

or 1.3 × 10–9 = (x + y)y .......(2)

Dividing equation (1) and (2), we get

$${x \over y} = 3.6$$

$$ \therefore $$ x = 3.6y

Putting in equation (2) we get

y(3.6y + y) = 1.3 × 10–9

$$ \Rightarrow $$ y = 1.68 $$ \times $$ 10-5

and x = 3.6 $$ \times $$ 1.68 $$ \times $$ 10-5 = 6.048 $$ \times $$ 10-5

$$ \therefore $$ [Ca2+] = (x + y) = (6.048 $$ \times $$ 10-5) + (1.68 $$ \times $$ 10-5)

$$ \therefore $$ [Ca2+] = 7.746 × 10–5 M

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