NEET Practice Questions (MCQs) with Answers & Solutions

$ if x is the solubility of AgCl in 0.04 M cacl2 , then $ [Ag ^ + ] = x mol L ^ {-1} $ $ [Cl ^ - ] = 2 \times 0.04 + x = 0.08 + x \cong 0.08 M $ $ K_{SP} of AgCl = [Ag ^ + ] [Cl ^ - ] $ $ 4 \times 10 ^ {- 10} / 0.08 = [Ag ^ + ] = 5 \times 10^ {-9} M $

$ P ^ H = pka + log { [ CH_3 COONa] \over [ CH_3 COOH] } $ $ 5.5 = 4.5 + log { [ CH_3 COONa ] \over 0.1 } $ $ 5.5 = 4.5 + log [ CH_3 COONa] + 1 $ $ \therefore log [ CH_3 COONa ] = 0 $ $ \therefore [ CH_3 CooNa] = 1 M $

$ I ^ - $ can accept protons and hence is a base.

According to the Lowry-Bronsted concept, an acid is a substance that can donate a proton (H⁺ ion). The $H_3O^+$ ion, also known as the hydronium ion, can donate a proton, hence it is considered an acid.

The conjugate acid of a base is formed when the base accepts a proton (H⁺ ion). For the base $NH_2^-$ (amide ion), accepting a proton forms $NH_3$ (ammonia), which is the conjugate acid of $NH_2^-$. Therefore, the conjugate acid of $NH_2^-$ is $NH_3$.

$ N_3 H \rightleftharpoons N_3 ^ - + H^ + Hydrazoic acid N_3 H $

In the reaction $NH_3 + Na ightarrow NaNH_2 + rac{1}{2} H_2$, ammonia ($NH_3$) donates a proton (H⁺) to form sodium amide ($NaNH_2$), acting as an acid according to the Lowry-Bronsted definition.

$ H_2S $ can donate proton but can't accept proton.

In the given reaction, $H_2SO_4$ donates a proton to $H_2O$, forming $HSO_4^-$ and $H_3O^+$. Therefore, $HSO_4^-$ is the conjugate base of $H_2SO_4$ and $H_2O$ is the conjugate base of $H_3O^+$.