Practice Start Test

Ionic Equilibrium

JEE Chemistry · 82 questions · Page 9 of 9 · Click an option or "Show Solution" to reveal answer

Q81
How many litres of water must be added to 1 litre of an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2?
A 0.1 L
B 0.9 L
C 2.0 L
D 9.0 L
Correct Answer
Option D
Solution

As

   pH=1;  H+=10−1=0.1 M\,\,\,pH = 1;\,\,{H^ + } = {10^{ - 1}} = 0.1\,M
   pH=2;  H+=10−2=0.01 M\,\,\,pH = 2;\,\,{H^ + } = {10^{ - 2}} = 0.01\,M

∴\therefore

   M1=0.1  V1=1;  \,\,\,{M_1} = 0.1\,\,{V_1} = 1;\,\,
M2=0.01  V2=?{M_2} = 0.01\,\,{V_2} = ?

From

M1V1=M2V2;0.1×1=0.01×V2{M_1}{V_1} = {M_2}{V_2};0.1 \times 1 = 0.01 \times {V_2}
V2=10 {V_2} = 10\,

litre ∴\therefore volume of water added

=10−1=9 = 10 - 1 = 9\,

litre.

Q82
The conjugate base of H2PO4- is :
A PO43−PO_4^{3-}
B HPO42−HPO_4^{2-}
C H3PO4
D P2O5
Correct Answer
Option B
Solution

NOTE : Conjugate acid-base differ by

H+{H^ + }
H2PO4−Acid→−H  HPO4−−conjugate  base\mathop {{H_2}PO_4^ - }\limits_{Acid} \mathop \to \limits^{ - H} \,\mathop {\,HPO{{_4^ - }^ - }}\limits_{conjugate\,\,base}
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