What is the ionic strength of 0.2M Na2HPO4 in a solution?
ionic strength = no. of ions produced by ionization , so in this case, it will be doubled
= 2*0.2= 0.4M
To calculate the ionic strength of a solution, we need to consider the concentration and charge of each ion present. The ionic strength, symbolized as I, is a measure of the total concentration of ions in a solution. It is calculated by summing the product of the concentration of each ion (in mol/L) and the square of its charge (in absolute units).
In the case of 0.2M Na2HPO4, let's break down the ions present:
- Na+ (Sodium ion) has a charge of +1
- HPO42- (Hydrogen phosphate ion) has a charge of -2
Since we have two sodium ions and one hydrogen phosphate ion, we need to consider those quantities in our calculation.
The concentration of Na+ ions will be 2 times the concentration of Na2HPO4, which is 0.2M x 2 = 0.4M.
The concentration of HPO42- ions will be equal to the concentration of Na2HPO4, which is 0.2M.
Now we can calculate the ionic strength:
I = (0.4 x (1^2)) + (0.2 x (-2^2))
I = (0.4 x 1) + (0.2 x 4)
I = 0.4 + 0.8
I = 1.2 mol/L
Therefore, the ionic strength of a 0.2M Na2HPO4 solution is 1.2 mol/L.