calculate total ionic strength of solution in which concentration of CuSO4 and Al2(SO4)3 is 0.1 molar
To calculate the total ionic strength of a solution, you need to determine the individual ionic strengths of each ion present in the solution and then sum them up.
The ionic strength (I) of a ion in a solution depends on its concentration (c) and charge (z). The equation to calculate ionic strength is:
I = (1/2) * Σ (c * z^2)
Let's calculate the ionic strengths for each ion in the solution:
For CuSO4:
- Cu^2+ ion has a charge of +2 and is present in a concentration of 0.1 M, so its ionic strength can be calculated as:
I(Cu^2+) = (1/2) * (0.1 * (2^2)) = 0.1
- SO4^2- ion has a charge of -2 and is present in a concentration of 0.1 M, so its ionic strength can be calculated as:
I(SO4^2-) = (1/2) * (0.1 * (2^2)) = 0.1
For Al2(SO4)3:
- Al^3+ ion has a charge of +3 and is present in a concentration of 0.1 M, so its ionic strength can be calculated as:
I(Al^3+) = (1/2) * (0.1 * (3^2)) = 0.45
- SO4^2- ion has a charge of -2 and is present in a concentration of 0.1 M, so its ionic strength can be calculated as:
I(SO4^2-) = (1/2) * (0.1 * (2^2)) = 0.1
Now, sum up all the ionic strengths to calculate the total ionic strength of the solution:
Total ionic strength = I(Cu^2+) + I(SO4^2-) + I(Al^3+) + I(SO4^2-)
= 0.1 + 0.1 + 0.45 + 0.1
= 0.75
Therefore, the total ionic strength of the solution containing 0.1 M CuSO4 and 0.1 M Al2(SO4)3 is 0.75.
CuSO4..two types of ions, .1M each
Al2(SO4)3 two types of ions, .2,and.3
total ionic strength: sum=.6M