Question

QUESTION 1 1.041 points If a 50.00 ml aliquot of a 0.12 M NaCl solution is added to 30.00 mL of a 0.18 M CaCl solution, what is the concentration of calcium ion in the mixture? 0.10M 0.086 M 0,068 M 0.36 M 0.090 M

Answer 1

[Ca²⁺] = 0.068 M

Step-by-step explanation:

The concentration of the calcium ion will be equal to the amount of calcium in CaCl₂, divided by the total volume:

C = n/V

When CaCl₂ dissociates in water, one mole of calcium ion is produced for every mole of CaCl₂, so the molar ratio of CaCl₂ to Ca²⁺ is 1:1. The moles of Ca²⁺ are calculated as follows:

(0.18 mol/L)(30.00 mL) = 5.4 mmol CaCl₂ = 5.4 mmol Ca²⁺

The total volume is (50.00 mL + 30.00 mL) = 80.00 mL

Thus, the concentration of Ca²⁺ is:

C = n/V = (5.4 mmol)/(80.00 mL) = 0.068 M