Question

Look at sample problem 19.10 in the 8th ed Silberberg book. Write the Ksp expression. Find the concentrations of the ions you need (in this case Ca2 and F-). Put those concentrations into the Ksp expression to calculate Q. Compare Q to K A common laboratory method for preparing a precipitate is to mix solutions containing the component ions. Does a precipitate form when 10. ml of 0.0010 M Ca(NO3)2 is mixed with 10. ml of 0.00010 M NaF

Answer 1

To predict precipitation when mixing solutions, calculate the ion product (Q) and compare it with the solubility product (Ksp). If Q exceeds Ksp, a precipitate will form. For the given problem, concentrations are calculated after mixing and then used to find Q to predict precipitation of calcium fluoride.

Step-by-step explanation:

When predicting whether a precipitate will form upon mixing two solutions, we must first determine the solubility product constant (Ksp) for the potential precipitate and the ion product (Q) for the mixture. The balanced equilibrium equation for calcium fluoride (CaF₂) is CaF₂ (s) ⇌ Ca²⁺ (aq) + 2F⁻ (aq), and the Ksp expression is Ksp = [Ca²⁺][F⁻]². In this scenario, after mixing equal volumes of 0.0010 M Ca(NO₃)₂ and 0.00010 M NaF, the final concentrations of Ca²⁺ and F⁻ need to be calculated considering dilution due to mixing.

To calculate Q, the resultant concentrations are inserted into the Ksp expression: Q = [Ca²⁺][F⁻]². If Q > Ksp, a precipitate will form. For the given concentrations of 0.0010 M Ca(NO₃)₂ and 0.00010 M NaF in a total volume of 20 mL, the concentrations after mixing are 0.00050 M Ca²⁺ and 0.000050 M F⁻. Therefore, Q = (0.00050)(0.000050)² = 1.25 × 10⁻¹ⁱ, which would be then compared to the Ksp of calcium fluoride (3.45 × 10⁻¹ⁱ) to determine precipitation.

Answer 2

Step-by-step explanation:

From the information given:

`CaF_2 \to Ca^(2+) + 2F^-`

`Ksp = 3.2 * 10^(-11)`

no of moles of
`Ca^(2+)` = 0.01 L × 0.0010 mol/L

no of moles of
`Ca^(2+)` =
`1 * 10^(-5) \ mol`

no of moles of
`F^-` = 0.01 L × 0.00010 mol/L

no of moles of
`F^-` =
`1 * 10^(-6)\ mol`

Total volume = 0.02 L

`[Ca^(2+)}] = (1*10^(-5) \ mol)/(0.02 \ L) \\ \\ \\ \[[Ca^(2+)}] = 0.0005 \ mol/L`

`[F^(-)] = ((1* 10^(-6) \ mol))/(0.02 \ L)`

`[F^(-)] = 5 * 10^(-5) \ mol/L`

`Q = [Ca^(2+)][F^-]^2 \\ \\ Q = 0.0005 * (5* 10^(-5))^2 \\ \\ Q = 1.25 * 10^(-12)`

Since Q<ksp, then there will no be any precipitation of CaF2