A25.00 ml solution of 0.150 m nacl is combined with 10.00 ml of a 0.0750 m cacl2 solution. assuming a total volume of 35.00 ml, determine the concentration of chloride ion in the combined solution.

[Cl⁻] = 1,5x10⁻⁴M

Explanation:

First of all, let's determinate the mole of each salt.

Molarity . volume = Mole

Volume must be in L, cause molarity is mol/L

NaCl  → Na⁺  + Cl⁻

Ratio is 1:1

0.15 mol/L . 0.025L = 3.75x10⁻³ mole

As ratio is 1:1, from 3.75x10⁻³ mole of salt, I have 3.75x10⁻³ mole of chloride

CaCl₂ → Ca²⁺  +  2Cl⁻

Ratio is 1:2 so, from 1 mol of salt I'll get the double of mole of chloride

0.075 mol/L . 0.010 L = 7.5x10⁻⁴ mol

7.5x10⁻⁴ mol . 2 = 1.5x10⁻³ mole

Total mole of Cl⁻:  3.75x10⁻³ +  1.5x10⁻³  = 5.25x10⁻³

This 5.25x10⁻³ mole are present in a total volume of 35 mL.

Let's convert 35 mL in L → 0.035L   (35/1000)

Molarity is mol/L → 5.25x10⁻³ mol / 0.035L = 1,5x10⁻⁴M

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