Physical Chemistry
Some Basic Concepts of Chemistry
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Basics of Organic Chemistry
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MCQ (Single Correct Answer)1
JEE Advanced 2022 Paper 1 Online
Numerical
+3
-0
A solution is prepared by mixing $0.01 \mathrm{~mol}$ each of $\mathrm{H}_{2} \mathrm{CO}_{3}, \mathrm{NaHCO}_{3}, \mathrm{Na}_{2} \mathrm{CO}_{3}$, and $\mathrm{NaOH}$ in $100 \mathrm{~mL}$ of water. $p \mathrm{H}$ of the resulting solution is _________.
[Given: $p \mathrm{~K}_{\mathrm{a} 1}$ and $p \mathrm{~K}_{\mathrm{a} 2}$ of $\mathrm{H}_{2} \mathrm{CO}_{3}$ are $6.37$ and 10.32, respectively; $\log 2=0.30$ ]
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2
JEE Advanced 2020 Paper 2 Offline
Numerical
+4
-0
A solution of 0.1 M weak base (B) is titrated with 0.1 M of a strong acid (HA). The variation of pH of the solution with the volume of HA added is shown in the figure below. What is the pKb of the base? The neutralisation reaction is given by
$$B + HA\buildrel {} \over \longrightarrow B{H^ + } + {A^ - }$$

$$B + HA\buildrel {} \over \longrightarrow B{H^ + } + {A^ - }$$

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3
JEE Advanced 2020 Paper 2 Offline
Numerical
+4
-0
An acidified solution of 0.05 M Zn2+ is saturated with 0.1 M H2S. What is the minimum molar concentration (M) of H+ required to prevent the precipitation of ZnS?
Use Ksp(ZnS) = 1.25 $$ \times $$ 10$$-$$22 and overall dissociation constant of
H2S, Knet = K1K2 = 1 $$ \times $$ 10-21.
Use Ksp(ZnS) = 1.25 $$ \times $$ 10$$-$$22 and overall dissociation constant of
H2S, Knet = K1K2 = 1 $$ \times $$ 10-21.
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4
JEE Advanced 2018 Paper 1 Offline
Numerical
+3
-0
The solubility of a salt of weak acid $$(AB)$$ at $$pH\,$$ $$3$$ is $$Y \times {10^{ - 3}}$$ $$mol\,{L^{ - 1}}.$$ The value of $$Y$$ is ________________.
(Given that the value of solubility product of $$AB$$ $$\left( {{K_{sp}}} \right) = 2 \times {10^{ - 10}}$$ and the value of ionization constant of $$HB$$ $$\left( {{K_a}} \right) = 1 \times {10^{ - 8}}$$)
(Given that the value of solubility product of $$AB$$ $$\left( {{K_{sp}}} \right) = 2 \times {10^{ - 10}}$$ and the value of ionization constant of $$HB$$ $$\left( {{K_a}} \right) = 1 \times {10^{ - 8}}$$)
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