The ionic strength of an aqueous solution containing 0.24 M CH3COOAg and 0.050 M NH4Cl is:
To calculate the ionic strength of a solution, you need to know the concentrations and charges of the ions present in the solution. In this case, the solution contains two ions: CH3COO- (acetate) and Ag+.
First, we need to find the concentration of each ion. The concentration of CH3COO- is given as 0.24 M. However, CH3COOAg dissociates fully in water, so the concentration of CH3COO- is equal to the concentration of CH3COOAg.
Next, we need to find the concentration of Ag+. Since CH3COOAg dissociates fully in water, the concentration of Ag+ is also 0.24 M.
The ionic strength (I) of the solution is calculated using the formula:
I = ∑ (1/2) * ci * zi^2
where ∑ represents the sum over all ions in the solution, ci is the molar concentration of the ion, and zi is the charge of the ion.
For this solution, we have two ions contributing to the ionic strength, CH3COO- and Ag+.
For CH3COO-:
I(CH3COO-) = (1/2) * 0.24 M * (-1)^2 = 0.12 M
For Ag+:
I(Ag+) = (1/2) * 0.24 M * (+1)^2 = 0.12 M
Finally, we add the ionic strength contributions from each ion to get the total ionic strength of the solution:
Total ionic strength = I(CH3COO-) + I(Ag+)
= 0.12 M + 0.12 M
= 0.24 M
Therefore, the ionic strength of the solution containing 0.24 M CH3COOAg and 0.050 M NH4Cl is 0.24 M.