Respuesta :
The question is incomplete, here is the complete question:
15.00 mL of a 0.127 M [tex]CaCl_2[/tex] solution are mixed with 25.60 mL of a 0.137 M [tex]CaCl_2[/tex] 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.
Explanation:
To calculate the molarity of the solution after mixing 2 solutions, we use the equation:
[tex]M=\frac{M_1V_1+M_2V_2}{V_1+V_2}[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of the first [tex]CaCl_2[/tex] solution
[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of the second [tex]CaCl_2[/tex] solution
We are given:
[tex]M_1=0.127M\\V_1=15.00mL\\M_2=0.137M\\V_2=25.60mL[/tex]
Putting all the values in above equation, we get:
[tex]M=\frac{(0.127\times 15.00)+(0.137\times 25.60)}{15.00+25.60}\\\\M=0.133M[/tex]
The chemical equation for the ionization of calcium chloride follows:
[tex]CaCl_2\rightarrow Ca^{2+}+2Cl^-[/tex]
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 = [tex](2\times 0.133)=0.266M[/tex]
Hence, the concentration of calcium and chloride ions in the resulting solution are 0.133 M and 0.266 M respectively.
