1

### 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.
2

### 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
3

### 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
4

### NEET 2013 (Karnataka)

At 100oC the Kw of water is 55 times its value at 25oC. What will be the pH of neutral solution? (log 55 = 1.74)
A
7.00
B
7.87
C
5.13
D
6.13

## Explanation

Kw at 25oC = 1 × 10–14

At 25ºC

Kw = [H+] [OH] = 10–14

At 100°C (given)

Kw = [H+] [OH] = 55 × 10–14

for a neutral solution

[H+] = [OH]

[H+]2 = 55 × 10–14

or [H+] = (55 × 10–14)1/2

pH = – log [H+]

On taking log on both side

– log [H+] = –log (55 × 10–14)1/2

$\Rightarrow$ pH = ${1 \over 2}$ - log55 - 14log10

$\Rightarrow$ pH = 6.13