1
BITSAT 2023
MCQ (Single Correct Answer)
+3
-1

If the ionic product of $$\mathrm{Ni}(\mathrm{OH})_2$$ is $$1.9 \times 10^{-15}$$, then the molar solubility of $$\mathrm{Ni}(\mathrm{OH})_2$$ in $$1.0 \mathrm{~M} \mathrm{~NaOH}$$ is

A
$$2.9 \times 10^{-18} \mathrm{~M}$$
B
$$1.9 \times 10^{-13} \mathrm{~M}$$
C
$$1.9 \times 10^{-15} \mathrm{~M}$$
D
$$2.9 \times 10^{-14} \mathrm{~M}$$
2
BITSAT 2022
MCQ (Single Correct Answer)
+3
-1

100 mL of 2 M of formic acid (pKa = 3.74) is neutralise by NaOH, at the equivalence point pH is

A
7
B
6
C
9.5
D
8.87
3
BITSAT 2021
MCQ (Single Correct Answer)
+3
-1

The solubility of Pb(OH)2 in water is 6.7 $$\times$$ 10$$-$$6 M. Its solubility in a buffer solution of pH = 8 would be :

A
1.2 $$\times$$ 10$$-$$2
B
1.6 $$\times$$ 10$$-$$3
C
1.6 $$\times$$ 10$$-$$2
D
1.2 $$\times$$ 10$$-$$3
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