Final answer:
The concentration of Sr₂+ when SrCrO₄ starts to precipitate from a solution that is 0.0025 M in CrO4₂- is 6.25 × 10⁻⁵ M. Thus the correct option is (d).
Explanation:
In this chemical reaction, SrCrO₄ is precipitating from a solution that initially has a concentration of 0.0025 M in CrO₄₂-. This process is called precipitation, which occurs when two soluble compounds react to form an insoluble compound. In this case, Sr₂+ and CrO₄₂- react to form SrCrO₄, which is insoluble in water. To calculate the concentration of Sr₂+ when precipitation occurs, we need to use the law of conservation of mass. This law states that the mass of a substance before and after a chemical reaction is the same. In other words, the amount of Sr₂+ before and after the reaction must be equal. Thus the correct option is (d).
Let's write the chemical equation for this reaction:
Sr₂+ (aq) + CrO₄₂- (aq) → SrCrO₄ (s)
We can see that one molecule of SrCrO₄ contains one molecule of Sr₂+ and one molecule of CrO₄₂-. Therefore, we can say that the ratio of Sr₂+ to CrO₄₂- in the solution is equal to the ratio of SrCrO₄ in the solid phase to CrO₄₂- in the solution. At this point, we know that the initial concentration of CrO4₂- is 0.0025 M, and we want to find the concentration of Sr₂+ when precipitation occurs. We can set up an ICE table to help us solve this problem:
| | Initial | Change | Final |
| Sr₂+ (M) | x | -x | x |
| CrO₄₂- (M) | 0.0025 | -x | (x) |
| Solid (g/L) | 0 | +x | x |
The initial concentration of Sr₂+ is unknown, so we'll call it x. When precipitation occurs, the solid phase will contain x grams per liter (g/L). Since we're dealing with molar concentrations, we need to convert grams per liter to moles per liter using Avogadro's number (6.02 × 10²³ mol/g). We can do this by multiplying the mass in grams by Avogadro's number and dividing by the molar mass of the compound:
Mass (g) = Molar mass (g/mol) × Moles (mol)
Moles (mol) = Mass (g) / Molar mass (g/mol)
Molarity (M) = Moles (mol/L) / Liter volume (L)
Using these conversions, we can calculate the final concentration of SrCrO₄ : Final concentration = Mass / Liter volume × Avogadro's number / Molar mass = x / x × 6.02 × 10²³ / 317 = 6.25 × 10⁻⁵ M
In summary, when precipitation occurs, the concentration of SrCrO₄ is equal to the initial concentration of CrO4₂-, which is 0.0025 M in this case. Since one molecule of SrCrO4 contains one molecule of Sr₂+ and one molecule of CrO4₂-, we can say that the concentration of Sr₂+ when precipitation occurs is also 6.25 × 10⁻⁵ M.