Question

SL 1000 MET 1 1 EA XT 2. An intravenous solution contains 4.0 x 10 mEq/L of Cland 15 mEq/L of HPO4. If Na' is the only cation in the solution, what is the Na' concentration, in mEq/L. 4.0 x 10 mEgaL 4.6x10 maal 1 m tq Nat - IL I ne tact - 4.0x10 nitq Nat 2 min Not 15 mEd HPO42- x is mta tena- Inta HP Q2 - - 30 mEq Net IL 3.0 x 10' +4.0x10'- 7,0x10 m Ea.

Answer 1

Answer:-

Given:-

concentration of Cl^- ion in solution = 4.0 $\times$ 10¹ mEq / L = 40  mEq / L

concentration of HPO₄^2- ion in solution = 15 mEq / L

total concentration of Na⁺ ion in solution = ?

As we know that Na⁺ is the only cation present in the solution.

therefore if

concentration of Cl^- ion in solution = 40  mEq / L

then

concentration of Na⁺ ion in solution =  40  mEq / L

because Na⁺ ion combined with Cl^- ion to form NaCl salt which given as follows:-

Na⁺ + Cl^-$\rightarrow$ NaCl

40 mEq / L 40 mEq / L 40 mEq / L

concentration of Na⁺ ion in solution that combined with Cl^- ion = 40 mEq / L

So 40  mEq / L Na⁺ ion combined with Cl^- ion in the solution.

similarly

if concentration of HPO₄^2- ion in solution = 15 mEq / L

then Na⁺ ion combined with HPO₄^2- ion to form Na₂ HPO₄  salt which given as follows:-

2Na⁺ + HPO₄^2-$\rightarrow$ Na₂ HPO₄

2 $\times$ 15 mEq / L 15 mEq / L 15 mEq / L

30 mEq / L 15 mEq / L 15 mEq / L

concentration of Na⁺ ion in solution that combined with HPO₄^2- ion = 30 mEq / L

So 30 mEq / L Na⁺ ion combined with HPO₄^2- ion in the solution.

therefore

total concentration of Na⁺ ion in solution = concentration of Na⁺ ion in solution that combined with Cl^- ion + concentration of Na⁺ ion in solution that combined with HPO₄^2- ion

total concentration of Na⁺ ion in solution = 40 mEq / L + 30 mEq / L

total concentration of Na⁺ ion in solution = 70 mEq / L (i.e the answer)

or

total concentration of Na⁺ ion in solution = 7.0 $\times$ 10¹ mEq / L (i.e the answer)