In this example given above, I am confused how to get the values from the Davies

Question

In this example given above, I am confused how to get the values from the Davies equation. Please demonstate the step by step solutions in getting the values from the Davies equation.

K, 1.75 X 10-5 mol/kg for acetic acid (HC2H302) in water at 25°C. Find the H3O+ and oH molalities in a 0.200-mol/kg 25°C aqueous solution of acetic acid. To solve (11.15) for m(H3O+), we need yt. To use the Davies equation (10.68) to estimate Yt, we need the ionic strength Im, which can't be calculated until m(H3Ot) is known. The solution to this dilemma is to first estimate m(H3ot) and mCK by setting y 1 in (11.15) and solving for the ionic mo- lalities. With these approximate molalities, we calculate an approximate Im and then use the Davies equation to find an approximate Yt, which we use in (11.15) to find a more accurate value for the molalities. If necessary, we can then use these more accurate molalities to find a more accurate La, and so on.

Explanation / Answer

First consider the following reaction : HX ---> H+ + X-

Initial m 0 0

Eqb. m-x x x   

Here initial denotes the initial molalities and eqb denotes the molalities at eqilibrium

Dissociation constant Ka = m(H+)*m(X-) / m(HX) = x*x / (m-x)

This is what has been shown above when we do not take into account the activity coefficients.

When activity coefficients are taken into account the equation becomes :

Ka = m(H+)*m(X-)*aH+*aX- / m(HX)

Here aH+ denotes activity for H+ ion

aH+ = fH+ * m(H+)

Here fH+ denotes activity coefficient, and m(H+) denotes molal conc.

fH+ is found using Davies equation, which in turn requires Ionic strength.

Thus we are stuck here in a situation which is like this :

We need to find A(molal conc.) which depends on B(activity coefficient). To calculate B (using Davies eqn)we need C(Ionic strength), but C depends on A.

Here we cannot find exact value of A since C also ultimately depends on A.

Thus we assume a value for B, which is its standard value ( 1 ) and then find A. We know it is incorrect to do so, but we still go with it and now after finding this incorrect A we use it to find C.

Now we have finally got the C value, we now use it to calculate B again. It should come as 1 because we assumed it to be 1 in the beginning, but it does not come equal to 1. This is the loop hole. Davies equation will not give exact value of 1 , rather some value less than 1. And this should also be the case because in real situations activity coefficients are not 1, they are 1 only in ideal cases.

Thus this gives us a value which is more realistic.

-log f = 0.5z1z2 ( I0.5/(1+I0.5) - 0.30*I )

z1 and z2 denote the charge on ions, in our case they are 1 and -1.

I is ionic strength and f is activity coefficient.

The values can be solved easily from this equation.

Let me know if you have any other difficulty.

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