Question

15.00 mL of a .127 M CaCl2 solution are mixed

with25.60 mL of a .137 M
Ca(NO3)2 solution. What is the
molarconcentration ion in the resulting solution?

Answer 1

The question is incomplete, here is the complete question:

15.00 mL of a 0.127 M
`CaCl_2` solution are mixed with 25.60 mL of a 0.137 M
`CaCl_2` solution. What is the molar concentration of ions in the resulting solution?

Answer: The concentration of calcium and chloride ions in the resulting solution are 0.133 M and 0.266 M respectively.

Step-by-step explanation:

To calculate the molarity of the solution after mixing 2 solutions, we use the equation:

`M=(M_1V_1+M_2V_2)/(V_1+V_2)`

where,

`M_1\text{ and }V_1` are the molarity and volume of the first
`CaCl_2` solution

`M_2\text{ and }V_2` are the molarity and volume of the second
`CaCl_2` solution

We are given:

`M_1=0.127M\\V_1=15.00mL\\M_2=0.137M\\V_2=25.60mL`

Putting all the values in above equation, we get:

`M=((0.127* 15.00)+(0.137* 25.60))/(15.00+25.60)\\\\M=0.133M`

The chemical equation for the ionization of calcium chloride follows:

`CaCl_2\rightarrow Ca^(2+)+2Cl^-`

1 mole of calcium chloride produces 1 mole of calcium ions and 2 moles of chloride ions

So, concentration of calcium ion = 0.133 M

Concentration of chloride ions =
`(2* 0.133)=0.266M`

Hence, the concentration of calcium and chloride ions in the resulting solution are 0.133 M and 0.266 M respectively.