Salt hydrolysis Questions in English

(D) $CH_3COONa$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$.
For the hydrolysis of a salt of a weak acid and a strong base, the $pH$ is calculated using the formula:
$pH = \frac{1}{2} (pK_w + pK_a + \log C)$
Expanding this, we get:
$pH = \frac{1}{2} pK_w + \frac{1}{2} pK_a + \frac{1}{2} \log C$
Therefore, the correct option is $D$.

(C) Ammonium acetate $(CH_3COONH_4)$ is a salt formed from a weak acid $(CH_3COOH)$ and a weak base $(NH_4OH)$.
For a salt of a weak acid and a weak base, the pH of the solution is given by the formula:
$pH = 7 + \frac{1}{2} pK_a - \frac{1}{2} pK_b$
Since the pH of such a salt solution is independent of the concentration (dilution), the pH remains constant upon dilution.
Given that $pK_a \approx pK_b$, the equation simplifies to:
$pH = 7 + \frac{1}{2} (pK_a - pK_b) = 7 + 0 = 7$
Therefore, the pH of the resulting solution is $7$.

(C) $1$. $CuSO_4$ is a salt of a strong acid $(H_2SO_4)$ and a weak base $(Cu(OH)_2)$.
$2$. In water, it dissociates into $Cu^{2+}$ and $SO_4^{2-}$ ions.
$3$. The $Cu^{2+}$ ion undergoes cationic hydrolysis: $Cu^{2+} + 2H_2O \rightleftharpoons Cu(OH)_2 + 2H^+$.
$4$. Due to the production of $H^+$ ions, the solution becomes acidic, resulting in a $pH < 7$.

(A) $Na_2CO_3$ is a salt of a strong base $(NaOH)$ and a weak acid $(H_2CO_3)$.
When dissolved in water, it undergoes anionic hydrolysis: $CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-$.
This reaction produces $OH^-$ ions, which increases the concentration of $[OH^-]$ and makes the solution basic $(pH > 7)$.

(D) $1$. $KNO_3$ is a salt of a strong acid $(HNO_3)$ and a strong base $(KOH)$, so its aqueous solution is neutral $(pH = 7)$.
$2$. $Na_2CO_3$ is a salt of a weak acid $(H_2CO_3)$ and a strong base $(NaOH)$, so its aqueous solution is basic $(pH > 7)$.
$3$. $CH_3COONa$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$, so its aqueous solution is basic $(pH > 7)$.
$4$. $CuCl_2$ is a salt of a strong acid $(HCl)$ and a weak base $(Cu(OH)_2)$. Due to the hydrolysis of the $Cu^{2+}$ ion, the solution becomes acidic $(pH < 7)$.

(C) Step $1$: Identify the acid and base components for each salt.
Step $2$: $Ammonium \ acetate$ is formed from $CH_3COOH$ (weak acid) and $NH_4OH$ (weak base).
Step $3$: $Sodium \ acetate$ is formed from $CH_3COOH$ (weak acid) and $NaOH$ (strong base).
Step $4$: $Copper(II) \ chloride$ $(CuCl_2)$ is formed from $HCl$ (strong acid) and $Cu(OH)_2$ (weak base).
Step $5$: $Sodium \ carbonate$ is formed from $H_2CO_3$ (weak acid) and $NaOH$ (strong base).
Step $6$: Therefore, $CuCl_2$ is the correct example of a salt of a strong acid and a weak base.

(B) $1$. Anionic hydrolysis occurs when the anion of a salt reacts with water to form a basic solution.
$2$. This happens in salts formed from a weak acid and a strong base.
$3$. $Na_2CO_3$ is a salt of a weak acid $(H_2CO_3)$ and a strong base $(NaOH)$.
$4$. The anion $CO_3^{2-}$ reacts with water: $CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-$, resulting in anionic hydrolysis.
$5$. $CuSO_4$, $NH_4Cl$, and $FeCl_3$ are salts of strong acids and weak bases, which undergo cationic hydrolysis.

(D) Step $1$: Hydrolysis occurs when a salt is formed from a weak acid and a weak base, a weak acid and a strong base, or a strong acid and a weak base.
Step $2$: $NaNO_3$ is a salt formed from a strong acid $(HNO_3)$ and a strong base $(NaOH)$.
Step $3$: Salts of strong acids and strong bases do not undergo hydrolysis because their ions do not react with water to change the $pH$ of the solution.
Step $4$: Therefore, $NaNO_3$ does not undergo hydrolysis.

(B) $(1)$ $Na_2CO_3$ is a salt of a strong base $(NaOH)$ and a weak acid $(H_2CO_3)$, so it is basic $(pH > 7)$.
$(2)$ $CuCl_2$ is a salt of a weak base $(Cu(OH)_2)$ and a strong acid $(HCl)$, so it is acidic $(pH < 7)$.
$(3)$ $CH_3COONa$ is a salt of a strong base $(NaOH)$ and a weak acid $(CH_3COOH)$, so it is basic $(pH > 7)$.
$(4)$ $NaNO_3$ is a salt of a strong base $(NaOH)$ and a strong acid $(HNO_3)$, so it is neutral $(pH = 7)$.
Therefore, $CuCl_2$ has the least $pH$ value.