Question

Serum samples are frequently analyzed for their Na+ content. A NaCl standard solution was prepared by carefully dissolving 1.2155 g in a 10.00 mL volumetric flask. Nest, 5.00 mL of this stock was used to "spike" a 95.00 mL serum sample. This spiked sample gave a signal height of 7.98. A pure serum sample gave a signal height of 4.27. A) What is the final concentration of the added standard in the spiked sample? B) Use answer A to find the original concentration of Na+ in the pure serum sample?

Answer 1

A) Mass of NaCl added as “spike” = 1.2155 g.

Molar mass of NaCl = (1*22.9897 + 1*35.453) g/mol = 58.4427 g/mol.

Volume of the “spike” sample prepared = 10.00 mL.

Molar concentration of NaCl “spike” prepared = (moles of NaCl)/(volume of spike in L) = [(1.2155 g)/(58.4427 g/mol)]/[(10.00 mL)*(1 L/1000 mL)] = (0.020798 mole)/(0.01 L) = 2.0798 mol/L ≈ 2.08 M (1 mol/L = 1 M) (ans).

Concentration of NaCl in g/mL = (1.2155 g)/(10.00 mL) = 0.12155 g/mL (ans) (this is the concentration of the standard added; the question doesn’t ask clearly whether you are looking for the concentration of the standard NaCl or of the standard Na⁺. Anywa, we will find the Na⁺ concentration in the spike below.).

B) In case of blood serum, working with g/mL concentrations is more predominant that with molar concentrations.

Let the serum sample originally contain x g/mL Na⁺. Assume that the signal intensity is directly proportional to the concentration of Na⁺. Hence, we have, for the original serum sample,

4.27 = K*x where K is the proportionality constant …..(1).

Consider the ionization of NaCl as below.

NaCl -------> Na⁺ + Cl^-

As per the stoichiometric equation,

1 mole NaCl = 1 mole Na⁺.

Therefore, 2.08 M NaCl = 2.08 M Na⁺ = (2.08 M)*(1 mol/L/1 M)*(22.9897 g/1 mole) Na⁺ = 47.818576 g/L Na⁺ = (47.818576 g)/[(1 L)*(1000 mL/1 L)] = 0.047818576 g/mL ≈ 0.0478 g/mL (atomic mass of Na = 22.9897 g/mol). Therefore, the NaCl “spike” has a concentration of 0.0478 g/mL Na⁺.

Mass of Na⁺ in the “spiked” sample = [(5.00 mL)*(0.0478 g/mL) + (95.00 mL)*(x g/mL)] = (0.239 + 95x) g.

The total volume of the “spiked” sample = (5.00 + 95.00) mL = 100.00 mL and the concentration of Na⁺ in the “spiked” sample = (0.239 + 95x) g/(100 mL) = (0.239 + 95x)/100 g/mL.

Again, assuming the signal intensity to be dependent only on the concentration of Na⁺ , we have,

7.98 = K*(0.239 + 95x)/100 ……(2)

Divide (2) by (1) and get

7.98/4.27 = [(0.239 + 95x)/100]/x = (0.239 + 95x)/(100x)

===> 1.868852 = (0.239 + 95x)/(100x)

===> 186.8852x = 0.239 + 95x

===> 91.8852x = 0.239

===> x = 0.239/91.8852 = 0.002601 ≈ 0.0026

The concentration of Na⁺ in the original serum sample is 0.0026 g/mL (ans).