View Avallable Hint(s)
Consider the dissociation of strong versus weak acids
$$
\begin{array}{l}
\mathrm{HCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) \
\mathrm{HF}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{I})=\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{F}^{-}(\mathrm{aq})
\end{array}
$$

The first reaction is not reversible, but the second one is. So, only the conjugate of the weak acid $\mathrm{F}^{\text {. }}$, can react with $\mathrm{H}_{3} \mathrm{O}^{+}$Now consider the solubility of insoluble salts
$$
\begin{array}{l}
\mathrm{AgCl}(\mathrm{s})=\mathrm{Ag}^{\prime}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) \
\mathrm{AgF}(\mathrm{s})=\mathrm{Ag}^{\prime}(\mathrm{aq})+\mathrm{F}^{-}(\mathrm{aq})
\end{array}
$$

The addition of acid has no effect on saver chloride. But for the second reaction, $\mathrm{H}_{3} \mathrm{O}^{4}$ will react with $\mathrm{F}$, decreasing is concentration and driving the equilibrium to the right. Thus, salts that contain the conjugate of a weak acid bocome more soluble as the acidity of the solution increases
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$\mathrm{Cs}(\mathrm{OH})_{2}$
$\mathrm{NIBr}_{2}$
$\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}$
$\mathrm{SrCO}_{3}$
$2 \mathrm{nS}^{2}$
$\mathrm{KI}$
Increase in solubity
No change in solubility

Question

View Avallable Hint(s)
Consider the dissociation of strong versus weak acids
$$
\begin{array}{l}
\mathrm{HCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) \
\mathrm{HF}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{I})=\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{F}^{-}(\mathrm{aq})
\end{array}
$$

The first reaction is not reversible, but the second one is. So, only the conjugate of the weak acid $\mathrm{F}^{\text {. }}$, can react with $\mathrm{H}_{3} \mathrm{O}^{+}$Now consider the solubility of insoluble salts
$$
\begin{array}{l}
\mathrm{AgCl}(\mathrm{s})=\mathrm{Ag}^{\prime}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) \
\mathrm{AgF}(\mathrm{s})=\mathrm{Ag}^{\prime}(\mathrm{aq})+\mathrm{F}^{-}(\mathrm{aq})
\end{array}
$$

The addition of acid has no effect on saver chloride. But for the second reaction, $\mathrm{H}_{3} \mathrm{O}^{4}$ will react with $\mathrm{F}$, decreasing is concentration and driving the equilibrium to the right. Thus, salts that contain the conjugate of a weak acid bocome more soluble as the acidity of the solution increases
Reset
Help
$\mathrm{Cs}(\mathrm{OH})_{2}$
$\mathrm{NIBr}_{2}$
$\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}$
$\mathrm{SrCO}_{3}$
$2 \mathrm{nS}^{2}$
$\mathrm{KI}$
Increase in solubity
No change in solubility

Accepted Answer

***C*** ∻Key Concept∻ ⚹Solubility and Acid-Base Reactions⚹ ∻Explanation∻ ⚹Salts containing the anions of weak acids increase in solubility in acidic solutions because the added $\mathrm{H_3O^+}$ reacts with the anion, reducing its concentration and shifting the equilibrium to dissolve more salt. $\mathrm{Ca_3(PO_4)_2}$ contains $\mathrm{PO_4^{3-}}$, the conjugate base of a weak acid, and thus its solubility will increase in acid.⚹◊Can you determine which salt will become more soluble with the addition of acid based on the given information?⍭ Generate me a similar question◊

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