Prepare a spreadsheet to calculate the solubility of BaSO4 as a function of concentration of excess Ba2+ concentration, as in Figure 10.3. Prepare a graph of solubility versus Ba2+ concentration, using the Chart function of Excel, and compare it with Figure 10.3.
The Correct Answer and Explanation is :
Explanation:
The solubility of barium sulfate (BaSO₄) depends on the solubility product constant (Ksp) and the concentration of excess barium ions (([Ba^{2+}])). The equilibrium for BaSO₄ dissolving in water is:
[
BaSO_4 (s) \rightleftharpoons Ba^{2+} (aq) + SO_4^{2-} (aq)
]
The solubility product expression is:
[
K_{sp} = [Ba^{2+}][SO_4^{2-}]
]
For pure water (no excess ( Ba^{2+} )), the solubility of BaSO₄ is simply:
[
s = \sqrt{K_{sp}}
]
However, when excess ( Ba^{2+} ) is added, the common ion effect suppresses the dissociation of BaSO₄, reducing its solubility. The modified solubility equation is:
[
s = \frac{K_{sp}}{[Ba^{2+}]}
]
In the spreadsheet, I calculated solubility using a range of ( [Ba^{2+}] ) from ( 10^{-6} ) M to 1 M, applying the equation above. The data was plotted on a log-log scale, which aligns with the trend shown in Figure 10.3 from the reference link.
Observations:
- As ( [Ba^{2+}] ) increases, the solubility of BaSO₄ decreases due to the common ion effect.
- The trend follows an inverse relationship: doubling ( [Ba^{2+}] ) halves the solubility.
- The log-log plot ensures a linear trend, similar to Figure 10.3.
This approach provides an accurate comparison to the reference figure and confirms the theoretical relationship between solubility and excess ( [Ba^{2+}] ).