Answer :
To find the concentration of [tex]\([Ba^{2+}]\)[/tex] in the intestinal tract, we need to calculate the solubility product constant ([tex]\(K_{sp}\)[/tex]) of the reaction where barium sulfate ([tex]\(BaSO_4\)[/tex]) dissolves slightly in water:
[tex]\[ BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq) \][/tex]
### Step-by-Step Solution
1. Given Information:
- [tex]\(\Delta G^{0} = 59.1 \, \text{kJ/mol}\)[/tex]
- Body temperature = [tex]\(37^{\circ} \text{Celsius}\)[/tex]
2. Convert [tex]\(\Delta G^{0}\)[/tex] to J/mol:
Since [tex]\(\Delta G^{0}\)[/tex] is given in kJ/mol, convert it to J/mol:
[tex]\[ \Delta G^{0} = 59.1 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 59100 \, \text{J/mol} \][/tex]
3. Convert Temperature to Kelvin:
Convert the body temperature from Celsius to Kelvin:
[tex]\[ T = 37 + 273.15 = 310.15 \, \text{K} \][/tex]
4. Calculate the Equilibrium Constant ([tex]\(K_{eq}\)[/tex]):
The relationship between [tex]\(\Delta G^{0}\)[/tex] and the equilibrium constant is given by:
[tex]\[ \Delta G^{0} = -RT\ln K_{eq} \][/tex]
Where [tex]\(R = 8.314 \, \text{J/(mol*K)}\)[/tex] is the universal gas constant.
Rearranging for [tex]\(K_{eq}\)[/tex],
[tex]\[ K_{eq} = \exp\left(-\frac{\Delta G^{0}}{RT}\right) \][/tex]
Substitute the values:
[tex]\[ K_{eq} = \exp\left(-\frac{59100}{8.314 \times 310.15}\right) \][/tex]
[tex]\[ K_{eq} = 1.112 \times 10^{-10} \][/tex]
5. Determine [tex]\([Ba^{2+}]\)[/tex]:
- For the dissolution of [tex]\(BaSO_4\)[/tex], the equilibrium relationship is:
[tex]\[ K_{sp} = [Ba^{2+}][SO_4^{2-}] \][/tex]
Since at equilibrium both [tex]\([Ba^{2+}]\)[/tex] and [tex]\([SO_4^{2-}]\)[/tex] are equal (due to the 1:1 stoichiometry of the dissolution),
[tex]\[ K_{sp} = [Ba^{2+}]^2 \][/tex]
Solving for [tex]\([Ba^{2+}]\)[/tex],
[tex]\[ [Ba^{2+}] = \sqrt{K_{sp}} \][/tex]
Substitute the value of [tex]\(K_{eq}\)[/tex] for [tex]\(K_{sp}\)[/tex]:
[tex]\[ [Ba^{2+}] = \sqrt{1.112 \times 10^{-10}} \][/tex]
[tex]\[ [Ba^{2+}] = 1.05 \times 10^{-5} \, \text{M} \][/tex]
Therefore, the concentration of [tex]\([Ba^{2+}]\)[/tex] in the intestinal tract is [tex]\(1.05 \times 10^{-5} \, \text{M}\)[/tex].
[tex]\[ BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq) \][/tex]
### Step-by-Step Solution
1. Given Information:
- [tex]\(\Delta G^{0} = 59.1 \, \text{kJ/mol}\)[/tex]
- Body temperature = [tex]\(37^{\circ} \text{Celsius}\)[/tex]
2. Convert [tex]\(\Delta G^{0}\)[/tex] to J/mol:
Since [tex]\(\Delta G^{0}\)[/tex] is given in kJ/mol, convert it to J/mol:
[tex]\[ \Delta G^{0} = 59.1 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 59100 \, \text{J/mol} \][/tex]
3. Convert Temperature to Kelvin:
Convert the body temperature from Celsius to Kelvin:
[tex]\[ T = 37 + 273.15 = 310.15 \, \text{K} \][/tex]
4. Calculate the Equilibrium Constant ([tex]\(K_{eq}\)[/tex]):
The relationship between [tex]\(\Delta G^{0}\)[/tex] and the equilibrium constant is given by:
[tex]\[ \Delta G^{0} = -RT\ln K_{eq} \][/tex]
Where [tex]\(R = 8.314 \, \text{J/(mol*K)}\)[/tex] is the universal gas constant.
Rearranging for [tex]\(K_{eq}\)[/tex],
[tex]\[ K_{eq} = \exp\left(-\frac{\Delta G^{0}}{RT}\right) \][/tex]
Substitute the values:
[tex]\[ K_{eq} = \exp\left(-\frac{59100}{8.314 \times 310.15}\right) \][/tex]
[tex]\[ K_{eq} = 1.112 \times 10^{-10} \][/tex]
5. Determine [tex]\([Ba^{2+}]\)[/tex]:
- For the dissolution of [tex]\(BaSO_4\)[/tex], the equilibrium relationship is:
[tex]\[ K_{sp} = [Ba^{2+}][SO_4^{2-}] \][/tex]
Since at equilibrium both [tex]\([Ba^{2+}]\)[/tex] and [tex]\([SO_4^{2-}]\)[/tex] are equal (due to the 1:1 stoichiometry of the dissolution),
[tex]\[ K_{sp} = [Ba^{2+}]^2 \][/tex]
Solving for [tex]\([Ba^{2+}]\)[/tex],
[tex]\[ [Ba^{2+}] = \sqrt{K_{sp}} \][/tex]
Substitute the value of [tex]\(K_{eq}\)[/tex] for [tex]\(K_{sp}\)[/tex]:
[tex]\[ [Ba^{2+}] = \sqrt{1.112 \times 10^{-10}} \][/tex]
[tex]\[ [Ba^{2+}] = 1.05 \times 10^{-5} \, \text{M} \][/tex]
Therefore, the concentration of [tex]\([Ba^{2+}]\)[/tex] in the intestinal tract is [tex]\(1.05 \times 10^{-5} \, \text{M}\)[/tex].
