Question

Example: Serum containing Na gave a signal of 4.27 mV in an atomic emission analysis. Then, 5 mL of 2.08 M NaCl were added to 95.0 mL of serum. The spiked serum gave a signal of 7.98 mV. Find the original concentration of Nat in serum.

Answer 1

Explanation:

It is known that,

       No. of moles = Concentration × volume

As 5 ml of 2.08 M NaCl NaS added to 95 ml of serum then,

        Total volume = (95 + 5) ml = 100 ml

Therefore, total moles of $Na^{+}$ = (5 × 2.08) + (95 × x)

where,      x =  concentration of $Na^{+}$ in original serum

Hence, signal obtained for this NaS = 7.8 mV

Let us assume same volume, that is, 100 ml Serum then $Na^{+}$ will be (100 × x) mol.

Signal obtained for this is 4.27 mV.

Now, the ratio of signal is equal to the ratio of mole.

So,   $\frac{4.27}{7.98} = \frac{100x}{(5 \times 2.08) + 95x}$

                           x = 0.113

Thus, we can conclude that original concentration of $Na^{+}$ in Serum is 0.113.