Mix Examples of Principles Related to Practical Chemistry Questions in English

(D) $I$ group basic radicals precipitate as chlorides. This is true,e.g.,$AgCl$,$PbCl_2$ (reagent is dil. $HCl$).
$(B)$ $IV$ group basic radicals precipitate as sulphides. This is true,e.g.,$ZnS$,$NiS$ (reagent used is $H_2S$ in ammoniacal medium).
$(C)$ $V$ group basic radicals precipitate as carbonates. This is true,e.g.,$BaCO_3$,$SrCO_3$ (reagent used is $(NH_4)_2CO_3$).
$\therefore$ Option $(D)$ is correct because all statements $(A)$,$(B)$,and $(C)$ are true.

(C) $1$. In the analysis of basic radicals,the group reagent for group-$IV$ is $H_2S$ in the presence of $NH_4OH$ and $NH_4Cl$. The statement in option $(A)$ is incomplete as it omits $H_2S$,but it is generally considered incorrect in the context of standard qualitative analysis reagents.
$2$. $dil. H_2SO_4$ is a non-oxidizing acid. It reacts with $NaBr$ to produce $HBr$ gas,which is colorless. Brown fumes of $Br_2$ are produced only when $conc. H_2SO_4$ (an oxidizing agent) is used. Thus,statement $(B)$ is incorrect.
$3$. Since both statements are scientifically inaccurate or incomplete,the correct choice is $(C)$.

(D) The reaction sequence is as follows:
$CaCO_3 (s) + 2CH_3COOH (aq) \rightarrow Ca(CH_3COO)_2 (aq) + H_2O (l) + CO_2 (g)$
In the first step,$CaCO_3$ reacts with acetic acid $(AcOH)$ to form soluble calcium acetate,releasing $CO_2$ gas.
In the second step,$Ca(CH_3COO)_2 (aq) + Na_2C_2O_4 (aq) \rightarrow CaC_2O_4 (s) + 2CH_3COONa (aq)$
Calcium oxalate $(CaC_2O_4)$ is a white precipitate.
Thus,$CO_2$ is evolved and both the initial substance $(CaCO_3)$ and the final precipitate $(CaC_2O_4)$ are white in colour.
Therefore,both observations $1$ and $3$ are correct.

(D) The reaction $BaCrO_4 \downarrow + CH_3COOH \text{ (Excess)} \longrightarrow \text{No reaction}$ indicates that the precipitate $BaCrO_4$ does not dissolve in the presence of excess acetic acid $(CH_3COOH)$.
This is because $CH_3COOH$ is a weak acid and cannot provide enough $H^+$ ions to shift the equilibrium of the dissolution reaction of $BaCrO_4$ significantly.
Therefore,this represents a case of 'No reaction'.

(A) The correct matches are:
$(i)$ Ammonium phosphomolybdate is $(NH_4)_3PO_4 \cdot 12MoO_3$ $(e)$.
$(ii)$ Ammonium phosphate is $(NH_4)_3PO_4$ $(d)$.
$(iii)$ Mixture of magnesia is $Mg^{2+} + NH_4OH$ $(b)$.
$(iv)$ Ammonium molybdate is $(NH_4)_2MoO_4$ $(a)$.
$(v)$ Phosphoric acid is $H_3PO_4$ $(c)$.
Thus,the correct sequence is $(i-e, ii-d, iii-b, iv-a, v-c)$.

$(A)$ In the kinetic study of the reaction between $I^-$ and $H_2O_2$, starch is used as an indicator for $I_2$.
Freshly prepared starch solution is required because old solutions may undergo bacterial degradation.
$Na_2S_2O_3$ is added to react with the produced $I_2$ to prevent the blue colour from appearing until all $Na_2S_2O_3$ is consumed.
Therefore, the concentration of $Na_2S_2O_3$ must be less than that of $KI$ to ensure the reaction proceeds and the blue colour appears.
The time is recorded immediately after the appearance of the blue colour, which indicates the consumption of $Na_2S_2O_3$.
Thus, statements $(A)$, $(B)$, and $(C)$ are correct.

(A) Prolonged heating is avoided during the preparation of ferrous ammonium sulphate because it causes the oxidation of $Fe^{2+}$ ions to $Fe^{3+}$ ions,which would contaminate the product.

(A) $Mg(NH_4)PO_4$ is a white precipitate formed during the test for magnesium ions.
$K_3[Co(NO_2)_6]$ (Potassium cobaltinitrite) is a yellow precipitate.
$MnO(OH)_2$ (or $MnO_2 \cdot H_2O$) is a brown precipitate.
$Fe_4[Fe(CN)_6]_3$ (Ferric ferrocyanide) is known as Prussian blue.
Therefore,the correct matching is $A-II, B-III, C-I, D-IV$.

(D) $SrSO_4$ is white.
$Mg(NH_4)PO_4$ is white.
$BaCrO_4$ is yellow.
$Mn(OH)_2$ is white.
$PbSO_4$ is white.
$PbCrO_4$ is yellow.
$AgBr$ is pale yellow.
$PbI_2$ is yellow.
$CaC_2O_4$ is white.
$[Fe(OH)_2(CH_3COO)]$ is brown-red.
The white coloured salts are $SrSO_4$,$Mg(NH_4)PO_4$,$Mn(OH)_2$,$PbSO_4$,and $CaC_2O_4$.
Total number of white salts = $5$.

(A, C) $NH_{4}NO_{3}$ reacts with conc. $NaOH$ to evolve pungent $NH_{3}$ gas.
It reacts with conc. $H_{2}SO_{4}$ to produce brown fumes of $NO_{2}$ (due to decomposition/redox).
It shows no visible reaction with water.
Similarly,$(NH_{4})_{2}CO_{3}$ reacts with conc. $NaOH$ to evolve pungent $NH_{3}$ gas.
It reacts with conc. $H_{2}SO_{4}$ to produce effervescence of $CO_{2}$ gas.
It shows no visible reaction with water.
Both $(a)$ and $(c)$ can distinguish the three beakers.

(B) Sodium dichromate $(Na_2Cr_2O_7)$ is highly hygroscopic,meaning it absorbs moisture from the atmosphere,which makes it unsuitable as a primary standard. Potassium dichromate $(K_2Cr_2O_7)$ is stable and non-hygroscopic,making it a primary standard. Thus,Statement $I$ is false.
Phenolphthalein is a weak organic acid,represented as $HIn$. It dissociates in a basic medium to form the colored $In^-$ ion. It does not act as a base,nor does it dissociate in an acidic medium. Thus,Statement $II$ is false.