Questions: mass of the resulting strontium fluoride precipitate. mass: 8 Assuming complete precipitation, calculate the final concentration of each ion. If the ion is no longer in solution, enter a 0 for the concentration. [Na+] = M [NO3] = M [St^2] = M [F-] = M

mass of the resulting strontium fluoride precipitate. mass: 8 Assuming complete precipitation, calculate the final concentration of each ion. If the ion is no longer in solution, enter a 0 for the concentration. [Na+] = M [NO3] = M [St^2] = M [F-] = M
Transcript text: mass of the resulting strontium flooride precipitate. mass: $\square$ 8 Assuming complete precipitation, calculate the final concentration of each ion. If the ion is no longer in solution, enter a 0 for the concentration. \[ \left[\mathrm{Na}^{+}\right]= \] $\square$ M \[ \left[\mathrm{NO}_{3}\right]= \] $\square$ M \[ \left[\mathrm{St}^{2}\right]= \] $\square$ M \[ [\mathbf{F}-\mathbf{]}= \] $\square$ M
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Solution

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Solution Steps

Step 1: Identify the Chemical Reaction

The reaction involves the precipitation of strontium fluoride (\(\text{SrF}_2\)) from a solution containing strontium ions (\(\text{Sr}^{2+}\)) and fluoride ions (\(\text{F}^-\)).

Step 2: Write the Balanced Chemical Equation

The balanced chemical equation for the precipitation reaction is: \[ \text{Sr}^{2+} + 2\text{F}^- \rightarrow \text{SrF}_2 \text{(s)} \]

Step 3: Determine the Initial Concentrations

Assume we have initial concentrations of \(\text{Na}^+\), \(\text{NO}_3^-\), \(\text{Sr}^{2+}\), and \(\text{F}^-\) ions. For simplicity, let's denote these initial concentrations as:

  • \([\text{Na}^+] = a\)
  • \([\text{NO}_3^-] = b\)
  • \([\text{Sr}^{2+}] = c\)
  • \([\text{F}^-] = d\)
Step 4: Calculate the Final Concentrations After Precipitation

Since \(\text{SrF}_2\) precipitates completely, the final concentration of \(\text{Sr}^{2+}\) and \(\text{F}^-\) will be zero. The concentrations of \(\text{Na}^+\) and \(\text{NO}_3^-\) will remain unchanged as they do not participate in the precipitation reaction.

Final Answer

\[ \left[\mathrm{Na}^{+}\right]= \boxed{a} \text{ M} \] \[ \left[\mathrm{NO}_{3}^{-}\right]= \boxed{b} \text{ M} \] \[ \left[\mathrm{Sr}^{2+}\right]= \boxed{0} \text{ M} \] \[ [\mathbf{F}^{-}]= \boxed{0} \text{ M} \]

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