# Sorum var244

## Hatice_Nur_Bayram said

The image you provided contains a chemistry problem related to solubility and the solubility product constant (K_{sp}). Let’s break down the question and find the answers.

### Question Breakdown:

• The solubility of a salt XY_2 is being observed at a certain temperature in a 10-liter aqueous solution.
• A graph is provided showing the change in concentration of X^{2+} ions over time as more XY_2 salt is added to the solution.
• There are three statements provided to determine their correctness based on the information given.

### Given Statements:

1. At the beginning, the concentration of X^{2+} ions in the solution is 3 \times 10^{-3} mol/L.

• This is a given value, we just need to confirm the initial condition from the graph.
2. At $t$°C, the solubility product constant, K_{sp}, for XY_2 is 3.2 \times 10^{-15}.

• This is to check the true value needed to determine the accuracy.
3. At $t$°C, the solubility of XY_2 salt is 2 \times 10^{-3} mol/L.

• This statement will be verified by the solubility product given.

### Solution Steps:

1. Check the beginning concentration:

• The initial concentration of X^{2+} ions from the graph looks like 3 \times 10^{-3} mol/L, so Statement I appears true.
2. Determine the solubility of XY_2:

• Solubility product expression:
K_{sp} = [X^{2+}][Y^-]^2
• Given K_{sp} = 3.2 \times 10^{-15}
• If the solubility (S) of XY_2 is 2 \times 10^{-3} mol/L, then:
[X^{2+}] = 2 \times 10^{-3} \text{ mol/L}
[Y^-] = 2S = 4 \times 10^{-3} \text{ mol/L}
• Calculate K_{sp}:
K_{sp} = (2 \times 10^{-3})(4 \times 10^{-3})^2 = 2 \times 10^{-3} \times 16 \times 10^{-6} = 32 \times 10^{-9} = 3.2 \times 10^{-8}
• The calculated K_{sp} does not match 3.2 \times 10^{-15}, so Statement III is incorrect.

### Correct Statements:

• I. Başlangıçta sulu çözeltide 3 \times 10^{-3} mol iyon bulunur.

### Sonuç:

Doğru ifadeler I ve II’dir.

C) I ve II.

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