- ASodium acetate and acetic acid in water
- BSodium acetate and hydrochloric acid in water
- CAmmonia and ammonium chloride in water
- ✓$(a) $ and $(c)$ both
Hint: A buffer solution can be prepared by mixing a weak acid or base with the salt of a strong base or strong acid.
Explanation:
Sodium acetate and acetic acid in water can form buffer solution because sodium acetate is a salt of strong acid $NaOH$ and weak acid acetic acid.
$CH _3 COONa + CH _3 COOH$
Sodium acetate and hydrochloric acid in water cannot form a buffer solution because $HCl$ is a strong acid.
Ammonia and ammonium chloride in water can from buffer solution because ammonia is a weak base and ammonium chloride is a salt of ammonia and strong acid $HCl$.
$NH _4 Cl + NH _3$
Ammonia and sodium hydroxide in water cannot form buffer solutions because both are bases.
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$\left( i \right)\,2F{e_2}{O_3}\left( s \right) \to 4Fe\left( s \right) + 3{O_2}\left( g \right)$
${\Delta _r}{G^o} = + 1487.0\,kJ\,mo{l^{ - 1}}$
$\left( {ii} \right)\,2CO\left( g \right) + {O_2}(g) \to 2C{O_2}\left( g \right)$
${\Delta _r}{G^o} = - 514.4\,kJ\,mo{l^{ - 1}}$
Free energy change, $\Delta_rG^o$ for the reaction
$\,2F{e_2}{O_3}\left( s \right) + 6CO\left( g \right) \to 4Fe\left( s \right) + 6C{O_2}\left( g \right)$ will be .....$kJ\, mol^{-1}$


$(1) Li < Be < B < C (IE_1)$
$(2) Li < Na < K < Rb < Cs$ (Reducing power in gaseous state)
$(3) Li^+ < Na^+ < K^+ < Rb^+ < Cs^+$ (Ionic mobility in aqueous solution)
$(4) S > Se > Te > O [EA]$