Question

We have a solution containing 0.10 M of NaCl and 0.05 M of CaCl2. The total concentration is [0.25,0.20,0.15] mEq/L for cation, and [0.15,0.20,0.25] mEq/L for anion. Please circle the correct answers.

Answer 1

NaCl ---> Na+ + Cl-

CaCl2 ----> Ca+2 + 2Cl-

For monovalent ions, 1 mEq = 1 mmol

For divalent ions, 1 mEq = 0.5 mmol

Considered 1L solution

Moles= molarity*V in L

Moles of Na+ = 0.1

Na ---> monovalent

10^-3 moles = 1 mEq

0.1 moles = 100 mEq

Ca---> is divalent

0.5 *10^-3 moles = 1 mEq

0.05 moles = 1*0.05 / 0.5*10^-3 = 100mEq

Total Cation concentration = 200 mEq/L= 0.2 Eq/L

Answer is 0.2 Eq/L not mEq/L ....units are wrong

For anion

Moles of Cl- from NaCl = 0.1

Moles of Cl- from CaCl2 = 0.05*2 = 0.1

Since Cl- is monovalent

Eq of Cl- = 0.1+0.1 = 0.2

Answer is 0.2 Eq/L not mEq/L ....units are wrong