Question

A student is carrying out a biological preparation that requires 1 M NaCl to maintain an ionic strength of 1.0. The student chooses to use 1.0 M ammonium sulfate instead. Why is this a serious error?

Answer 1

Explanation:

It is known that the ionic strength of a solution, I, is represented as follows.

          $I = \frac{1}{2} \sum m_{i}z^{2}_{i}$

where,  $m_{i}$ = the concentration of the ion,

             $z_{i}$ = the charge of the ion

Now, for 1 M NaCl solution, the ionic strength will be calculated as follows.

           I = $\frac{1}{2} [(1 \times 1^{2}) + (1 \times 1^{2})]$

             = 1.00

And, for 1 M $(NH_{4})_{2}SO_{4}$, concentration of $[NH_{4}]$ = 2 M and $[SO_{4}]$ = 1 M.

             I = $\frac{1}{2} [(2 \times 1^{2}) + ((1 \times 2^{2})]$

               = 3.00

As the ionic strength of 1 M $(NH_{4})_{2}SO_{4}$, is different  therefore it is a serious error.

Answer 2

Explanation:

The ionic strength of a solution, I, is defined as

I=1/2 Σ m_i z_i^2

Where m_i is the concentration of the ion, and z_i is the charge of the ion.

For 1 M NaCl solution, Ionic strength I = ½ (1*12 + 1*12) = 1.00;

For, 1 M (NH4)2SO4, concentration of [NH4] = 2 M and [SO4] = 1 M;

I = ½ (2*12 + 1*22) = 3.00;

so ionic strength of 1 M (NH4)2SO4 is different, therefore it is a serious error.