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AIEEE 2010 Solved Paper

Section: Chemistry

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Question : 10 of 89

Marks: +1, -0

Solubility product of silver bromide is

5.0 \times 10^{-13}

. The quantity of potassium bromide (molar mass taken as

120 \text{ g}

\text{mol}^{-1}

) to be added to 1 litre of

0.05 \text{ M}^{\;\text{Molution of silver }\;}

nitrate to start the precipitation of

\mathrm{AgBr}

$1.2 \times 10^{-10} \text{ g}$
$1.2 \times 10^{-9} \text{ g}$
$6.2 \times 10^{-5} \text{ g}$
$5.0 \times 10^{-8} \text{ g}$

Solution:

A g B r \rightleftharpoons A g^{+} + B r^{-}

K_{s p} = [A g^{+}] [B r^{-}]

For precipitation to occur

Ionic product

>

Solubility product

[B r^{-}] = \; \frac{K_{s p}}{[A g^{+}]} = \; \frac{5 \times 10^{-13}}{0.05} = 10^{-11}

i.e., precipitation just starts when

10^{-11}

moles of

K \mathrm{Br}

is added to

1 \ell \mathrm{AgNO} \mathrm{NO}_3

solution

\therefore

Number of moles of

\mathrm{Br}^{-}

needed

from

K B r = 10^{-11}

\; \therefore \;\; \; \text{ Mass of } \; K B r = 10^{-11} \times 120

\; = 1.2 \times 10^{-9} \text{ g}

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