Question

What concentration of clo3– results when 925 ml of 0.393 m agclo3 is mixed with 685 ml of 0.283 m mn(clo3)2?

Answer 1

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 2

Answer: The concentration of chlorate ion is 0.467 M

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$     .....(1)

• For $AgClO_3$:

Molarity of $AgClO_3$ 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:

$0.393M=\frac{\text{Moles of }AgClO_3}{0.925L}\\\\\text{Moles of }AgClO_3=(0.393mol/L\times 0.925L)=0.364mol$

• For $Mg(ClO_3)_2$:

Molarity of $Mg(ClO_3)_2$ solution = 0.283 M

Volume of solution = 685 mL = 0.685 L

Putting values in equation 1, we get:

$0.283M=\frac{\text{Moles of }Mg(ClO_3)_2}{0.685L}\\\\\text{Moles of }Mg(ClO_3)_2=(0.283mol/L\times 0.685L)=0.194mol$

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:

$Mg(ClO_3)_2\rightarrow Mg^{2+}+2ClO_3^-$

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:

$AgClO_3\rightarrow Ag^++ClO_3^-$0.752$\text{Molarity of chlorate ion}=\frac{0.752mol}{1.610L}\\\\\text{Molarity of chlorate ion}=0.467M$

Hence, the concentration of chlorate ion is 0.467 M