Respuesta :
Answer;
= 0.7698 M
Explanation and solution;
AgClO3 ionizes as follows:
AgClO3 ---> Ag^+ + ClO3^-
Moles AgClO3 dissolved ---> (0.393 mol/L) (0.925 L) = 0.424865 mol
From the chemical equation, one mole of AgClO3 dissolving yields one mole of ClO3^- in solution.
Moles ClO3^- = 0.424 865 mol
Similarly;
Mn(ClO3)2 dissolves as follows:
Mn(ClO3)2 ---> Mn^2+ + 2ClO3^-
Moles Mn(ClO3)2 dissolved ---> (0.283 mol/L) (0.685 L) = 0.413139 mol
From the chemical equation, one mole of Mn(ClO3)2 dissolving yields two moles of ClO3^- in solution.
Moles ClO3^- = 0.413139 mol x 2 = 0.826277 mol
Total moles ClO3^- in solution;
0.826277 mol + 0.413139 mol = 1.239416 mol
Total volume of solution ---> 0.925 L + 0.685 L = 1.61 L
Molarity of ClO3^- ---> 1.239416 mol / 1.61 L = 0.7698 M
Answer: The concentration of chlorate ion is 0.467 M
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex] .....(1)
- For [tex]AgClO_3[/tex]:
Molarity of [tex]AgClO_3[/tex] solution = 0.393 M
Volume of solution = 925 mL = 0.925 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:
[tex]0.393M=\frac{\text{Moles of }AgClO_3}{0.925L}\\\\\text{Moles of }AgClO_3=(0.393mol/L\times 0.925L)=0.364mol[/tex]
- For [tex]Mg(ClO_3)_2[/tex]:
Molarity of [tex]Mg(ClO_3)_2[/tex] solution = 0.283 M
Volume of solution = 685 mL = 0.685 L
Putting values in equation 1, we get:
[tex]0.283M=\frac{\text{Moles of }Mg(ClO_3)_2}{0.685L}\\\\\text{Moles of }Mg(ClO_3)_2=(0.283mol/L\times 0.685L)=0.194mol[/tex]
The chemical equation for ionization of silver chlorate follows:
1 mole of silver chlorate produces 1 mole of silver ion and 1 mole of chlorate ion
Moles of chlorate ion = 0.364 moles
The chemical equation for ionization of magnesium chlorate follows:
[tex]Mg(ClO_3)_2\rightarrow Mg^{2+}+2ClO_3^-[/tex]
1 mole of magnesium chlorate produces 1 mole of magnesium ion and 2 moles of chlorate ion
Moles of chlorate ion = (2 × 0.194) = 0.388 moles
- Now, calculating the molarity of chlorate ion by using equation 1, we get:
Moles of chlorate ion = (0.364 + 0.388) = 0.752 moles
Volume of solution = (925 + 685) = 1610 mL = 1.610 L
Putting values in equation 1, we get:
[tex]AgClO_3\rightarrow Ag^++ClO_3^-[/tex]0.752[tex]\text{Molarity of chlorate ion}=\frac{0.752mol}{1.610L}\\\\\text{Molarity of chlorate ion}=0.467M[/tex]
Hence, the concentration of chlorate ion is 0.467 M
