So in the past, we learned that our activity coefficient, which is represented by gamma, and ionic strength, which is represented by mu, of a solution could be closely and accurately related by using the extended Debye-Hückel equation. Here, we'd say that logγ1+α/1α×√μ305 . When the size parameter of the ion, which is alpha, is unknown, we're going to use instead the Davies equation. Here, because of the lack of a size parameter, this formula is most useful for monovalent ions. So ions with a charge of 1, plus or minus. Examples would be sodium or chloride ions. Now, you could also use larger charges, but there tends to be greater deviation from a credible value, but you can still use larger charged ions as well.

Here, the equation is reformatted into logγ=-0.51z2√μ1+√μ-0.3×μ. Here we have our ionic strength values. They're increasing as we go down. Here, we have our charges, plus or minus 1, plus or minus 2, plus or minus 3, and our activity coefficients. Notice that as your ionic strength is increasing, your activity coefficients are decreasing.

Now, we're going to say here from the Davies equations, all ions with the same magnitude in charge. So if 2 ions, one is plus 1 and the other one is minus 1 or plus 2 and minus 2, they will have the same activity coefficient, which is what we're seeing here. Plus or minus, it really doesn't matter. Again, that's because of the lack of an alpha value. We can group things together and make our list overall smaller. Traditionally, when we're using the extended Debye-Hückel equation, we segregate ions based on their charge as well as the value of their charge.

Now that we've seen this, attempt to do the practice question that's left here at the bottom of the page, where we're asked to determine what the activity coefficient is of calcium ion when we have 0.025 molar of calcium phosphate. Here, we don't have the size parameters, so we're not going to be able to use our typical extended Debye-Hückel equation. Try this question out. If you get stuck, don't worry. Come back and see how I answer the same question.