Question

compare the solubility of silver sulfite in each of the following aqueous solutions: 0.10 m agch₃coo, 0.10 m (nh₄)₂so₃, 0.10 m nh₄ch₃coo, 0.10 m nullo₃. options: more soluble than in pure water, similar solubility as in pure water, less soluble than in pure water.

Explanation:

Step1: Analyze Silver Sulfite Solubility

Silver sulfite (\(Ag_2SO_3\)) dissociates as \(Ag_2SO_3(s) ightleftharpoons 2Ag^+(aq) + SO_3^{2-}(aq)\). We use the common ion effect: if a solution has a common ion (\(Ag^+\) or \(SO_3^{2-}\)) with \(Ag_2SO_3\), solubility decreases; no common ion means solubility similar or more (if ion reacts).

Step2: Analyze 0.10 M \(AgCH_3COO\)

\(AgCH_3COO\) dissociates to \(Ag^+(aq) + CH_3COO^-(aq)\). Common ion \(Ag^+\) with \(Ag_2SO_3\). Common ion effect: equilibrium shifts left, solubility of \(Ag_2SO_3\) is **Less soluble than pure water**.

Step3: Analyze 0.10 M \((NH_4)_2SO_3\)

\((NH_4)_2SO_3\) dissociates to \(2NH_4^+(aq) + SO_3^{2-}(aq)\). Common ion \(SO_3^{2-}\) with \(Ag_2SO_3\). Common ion effect: equilibrium shifts left, solubility of \(Ag_2SO_3\) is **Less soluble than pure water**. Wait, no—wait, original problem: maybe I misread. Wait, silver sulfite: \(Ag_2SO_3\), so \(SO_3^{2-}\) is common. So \((NH_4)_2SO_3\) has \(SO_3^{2-}\), so solubility less. But wait, the options: let's recheck.

Wait, 0.10 M \(NaNO_3\): no common ions with \(Ag_2SO_3\) (ions \(Na^+\), \(NO_3^-\)). So solubility **Similar solubility as pure water** (since no common ion, and \(NaNO_3\) is a neutral salt, no hydrolysis affecting \(Ag^+\) or \(SO_3^{2-}\)).

0.10 M \(NH_4CH_3COO\): \(NH_4^+\) and \(CH_3COO^-\). \(NH_4^+\) hydrolyzes to \(H^+\), \(CH_3COO^-\) hydrolyzes to \(OH^-\), but no common ion with \(Ag_2SO_3\). Wait, no—wait, silver sulfite: \(Ag^+\) and \(SO_3^{2-}\). \(NH_4CH_3COO\) has no \(Ag^+\) or \(SO_3^{2-}\), so solubility similar? No, wait, maybe I made a mistake. Let's list each:

1. \(0.10\ M\ AgCH_3COO\): provides \(Ag^+\) (common ion), so solubility of \(Ag_2SO_3\) is **Less soluble than pure water** (common ion effect).

2. \(0.10\ M\ (NH_4)_2SO_3\): provides \(SO_3^{2-}\) (common ion), so solubility of \(Ag_2SO_3\) is **Less soluble than pure water**? Wait, no—wait the problem is to compare each solution:

Wait, the top boxes are the solutions: \(0.10\ M\ AgCH_3COO\), \(0.10\ M\ (NH_4)_2SO_3\), \(0.10\ M\ NH_4CH_3COO\), \(0.10\ M\ NaNO_3\). The bottom boxes are the solubility comparisons.

Let's correct:

- \(0.10\ M\ AgCH_3COO\): common ion \(Ag^+\) → solubility of \(Ag_2SO_3\) is **Less soluble than pure water**.

- \(0.10\ M\ (NH_4)_2SO_3\): common ion \(SO_3^{2-}\) → solubility of \(Ag_2SO_3\) is **Less soluble than pure water**? Wait, no—wait, maybe I messed up. Wait, silver sulfite: \(Ag_2SO_3\) dissociates into \(2Ag^+ + SO_3^{2-}\). So:

- \(AgCH_3COO\) gives \(Ag^+\) (common ion) → solubility decreases (less than pure water).

- \((NH_4)_2SO_3\) gives \(SO_3^{2-}\) (common ion) → solubility decreases (less than pure water).

- \(NH_4CH_3COO\): \(NH_4^+\) and \(CH_3COO^-\). No common ion. But \(NH_4^+\) hydrolyzes to \(H^+\), \(CH_3COO^-\) to \(OH^-\), but does that affect \(Ag_2SO_3\)? \(SO_3^{2-}\) is a base, so \(H^+\) from \(NH_4^+\) hydrolysis would react with \(SO_3^{2-}\): \(SO_3^{2-} + H^+ ightleftharpoons HSO_3^-\). This removes \(SO_3^{2-}\) from solution, shifting \(Ag_2SO_3\) dissociation right, so solubility **More soluble than pure water**? Wait, this is a complex case. But maybe the problem simplifies:

- \(NaNO_3\): no common ions, no hydrolysis affecting \(Ag^+\) or \(SO_3^{2-}\) → **Similar solubility as pure water**.

Let's re-express the correct matches (assuming standard common ion effect for simple cases, ignoring hydrolysis for some):

- \(0.10\ M\ AgCH_3COO\): common \(Ag^+\) → Less soluble than pure water.

- \(0.10\ M\ (NH_4)_2SO_3\): common \(SO_3^{2-}\) → Less soluble than pure water.

- \(0.10\ M\ NH_4CH_3COO\): if \(NH_4^+\) hydrolyzes \(H^+\) to react with \(SO_3^{2-}\), then solubility more. But maybe the problem considers \(NH_4CH_3COO\) as having no common ion, but hydrolysis: this is tricky. However, the key is:

- \(NaNO_3\): no common ion, neutral salt → Similar solubility as pure water.

So the matches (assuming the problem's intended approach, ignoring hydrolysis for \(NH_4CH_3COO\) as a simple case):

- \(0.10\ M\ AgCH_3COO\) → Less soluble than pure water.

- \(0.10\ M\ (NH_4)_2SO_3\) → Less soluble than pure water (common \(SO_3^{2-}\)).

- \(0.10\ M\ NH_4CH_3COO\) → More soluble than pure water (if \(H^+\) from \(NH_4^+\) hydrolysis reacts with \(SO_3^{2-}\), removing it, shifting equilibrium right).

- \(0.10\ M\ NaNO_3\) → Similar solubility as pure water.

But the problem's boxes: top four solutions, bottom three options. Wait, the image shows four top boxes: \(0.10\ M\ AgCH_3COO\), \(0.10\ M\ (NH_4)_2SO_3\), \(0.10\ M\ NH_4CH_3COO\), \(0.10\ M\ NaNO_3\). Bottom three options: "Less soluble...", "Similar...", "More soluble...". So we need to match each top to bottom.

Correct matches (standard common ion effect, and hydrolysis for \(NH_4CH_3COO\)):

1. \(0.10\ M\ AgCH_3COO\): common \(Ag^+\) → Less soluble than pure water.

2. \(0.10\ M\ (NH_4)_2SO_3\): common \(SO_3^{2-}\) → Less soluble than pure water. Wait, but there's only one "Less soluble..." box? No, the image shows three bottom boxes: "Less soluble...", "Similar...", "More soluble...". Wait, maybe the original problem has four top boxes and three bottom boxes, so one bottom box is used twice? Or maybe I misread the compounds.

Wait, the user's image: top boxes (left to right):

1. \(0.10\ M\ AgCH_3COO\)

2. \(0.10\ M\ (NH_4)_2SO_3\)

3. \(0.10\ M\ NH_4CH_3COO\)

4. \(0.10\ M\ NaNO_3\)

Bottom boxes (left to right):

1. Less soluble than pure water.

2. Similar solubility as pure water.

3. More soluble than pure water.

So we need to match each top to a bottom.

- \(0.10\ M\ AgCH_3COO\): common \(Ag^+\) → Less soluble (bottom 1).

- \(0.10\ M\ (NH_4)_2SO_3\): common \(SO_3^{2-}\) → Less soluble (bottom 1) – but there's only one "Less soluble" box? Wait, maybe the problem has a typo, or I made a mistake. Wait, silver sulfite: \(Ag_2SO_3\), so \(SO_3^{2-}\) is the anion. \((NH_4)_2SO_3\) provides \(SO_3^{2-}\), so common ion, solubility less.

- \(0.10\ M\ NH_4CH_3COO\): \(NH_4^+\) hydrolyzes to \(H^+\), \(CH_3COO^-\) hydrolyzes to \(OH^-\), but \(H^+\) can react with \(SO_3^{2-}\) (a base) to form \(HSO_3^-\), removing \(SO_3^{2-}\) from solution. This shifts the \(Ag_2SO_3\) dissociation equilibrium to the right (since \(SO_3^{2-}\) is removed), so solubility of \(Ag_2SO_3\) is **More soluble than pure water** (bottom 3).

- \(0.10\ M\ NaNO_3\): no common ions, neutral salt, no hydrolysis affecting \(Ag^+\) or \(SO_3^{2-}\) → **Similar solubility as pure water** (bottom 2).

So the matches are:

- \(0.10\ M\ AgCH_3COO\) → Less soluble than pure water.

- \(0.10\ M\ (NH_4)_2SO_3\) → Less soluble than pure water (but if the bottom box "Less soluble" is one, maybe the problem considers \((NH_4)_2SO_3\) as common ion, \(AgCH_3COO\) as common ion, \(NH_4CH_3COO\) as more, \(NaNO_3\) as similar.

Answer:

• \(0.10\ M\ AgCH_3COO\) → Less soluble than pure water.
• \(0.10\ M\ (NH_4)_2SO_3\) → Less soluble than pure water.
• \(0.10\ M\ NH_4CH_3COO\) → More soluble than pure water.
• \(0.10\ M\ NaNO_3\) → Similar solubility as pure water.

(Note: The exact matching depends on the problem's intended complexity, but the key is applying the common ion effect and hydrolysis for ion removal.)