Question

Indicate the concentration of each ion present in the solution formed by mixing the following. (Assume that the volumes are additive.)

3.50 g of NaCl in 49.5 mL of 0.551 M CaCl₂ solution
Na⁺

________M

Ca²⁺

_________M

Cl ^-

_________M

Answer 1

First calculate number of moles of NaCl

No. of moles = Mass given / Molar mass

Molar mass of NaCl (M) = 58.44

No. of moles of NaCl = 3.50 / 58.44 = 0.059 moles

0.059 moles of NaCl will 0.059 moles of Na⁺ and 0.059 moles of Cl^- ions

Molarity = No. of moles / Volume of solution (in L) = 0.059 / 0.0495 = 1.19 M

[Na+] = 1.19 M

Given: 49.5 mL of 0.551 M CaCl₂

0.551 M CaCl₂ on dissociation will give 0.551 M Ca+ ions and 1.102 M Cl^- ions. (Since one molecule of CaCl₂ gives one Ca2+ ion ans two Cl- ion)

[Ca²⁺] = 0.551 M

Number of moles of CaCl₂ = Volume (in L) x molarity = 49.5 x 10^-3 x 0.551 = 0.027 moles.

Number of moles of Cl^- ion = 0.027 x 2 = 0.0545 moles

Total number of moles of Cl- ion = 0.059 + 0.0545 = 0.1135 moles

[Cl-] = 0.1135 /0.0495 =2.29 M