Question

Consider the solubility product of barite (BaSO₄) at 25 °C and 1 bar to be 10-¹⁰. If the activities of \(Ba^{2+}\) and \(SO^{2-}_4\) ions are 0.5×10^-5 and 10^-X, respectively, then the absolute value of 'X' is ____(rounded off to one decimal place). 

Hint:

Always check unit consistency in solubility product problems and express values in proper scientific notation before applying logarithms. This avoids rounding errors.

Answer 1

Correct Answer: 4.6

Step 1: Write the solubility product expression \[ K_{sp} = [\text{Ba}^{2+}] \cdot [\text{SO}_4^{2-}] \] Step 2: Substitute known values \[ 10^{-10} = (0.5 \times 10^{-5}) \cdot (10^{-X}) \] Step 3: Rearrange for \(10^{-X}\) \[ 10^{-X} = \frac{10^{-10}}{0.5 \times 10^{-5}} \] Step 4: Simplify the denominator \[ 10^{-X} = \frac{10^{-10}}{5 \times 10^{-6}} = \frac{10^{-10}}{5 \cdot 10^{-6}} \] Step 5: Simplify further \[ 10^{-X} = \frac{1}{5} \times 10^{-4} \] Step 6: Express in standard form \[ 10^{-X} = 0.2 \times 10^{-4} = 2 \times 10^{-5} \] Step 7: Take log to solve for \(X\) \[ -\,X = \log_{10}(2 \times 10^{-5}) \] \[ -\,X = \log_{10} 2 + \log_{10} 10^{-5} \] \[ -\,X = 0.3010 - 5 \] \[ -\,X = -4.699 \] Step 8: Final result \[ X = 4.699 \approx 4.7 \] \[ \boxed{X \approx 4.7} \]