The acid dissociation constant, Ka
You can get a measure of the position of an equilibrium by writing an equilibrium constant for the reaction. The lower the value for the constant, the more the equilibrium lies to the left. The dissociation (ionization) of an acid is an example of a homogeneous reaction. Everything is present in the same phase – in this case, in solution in water. You can therefore write a simple expression for the equilibrium constant, Kc. Here is the equilibrium again:
[HA + H_2O rightleftharpoons H_3O^+ + A^- tag{5}]
You are viewing: Which Of The Following Describes A Strong Acid
You might expect the equilibrium constant to be written as:
[]K_c = dfrac{[H_3O^+][A^-]}{[HA][H_2O]} ]
However, if you think about this carefully, there is something odd about it.
At the bottom of the expression, you have a term for the concentration of the water in the solution. That’s not a problem – except that the number is going to be very large compared with all the other numbers.
Read more : Which Of The Following Is A Result Of Cost Distortion
In 1 dm3 of solution, there are going to be about 55 moles of water.
If you had a weak acid with a concentration of about 1 mol dm-3, and only about 1% of it reacted with the water, the number of moles of water is only going to fall by about 0.01. In other words, if the acid is weak the concentration of the water is virtually constant. In that case, there isn’t a lot of point in including it in the expression as if it were a variable. Instead, a new equilibrium constant is defined which leaves it out. This new equilibrium constant is called Ka.
You may find the Ka expression written differently if you work from the simplified version of the equilibrium reaction:
This may be written with or without state symbols.
It is actually exactly the same as the previous expression for Ka! Remember that although we often write H+ for hydrogen ions in solution, what we are actually talking about are hydroxonium ions. This second version of the Ka expression is not as precise as the first one.
To take a specific common example, the equilibrium for the dissociation of ethanoic acid is properly written as:
[CH_3COOH + H_2O rightleftharpoons CH_3COO^- + H_3O^+ tag{7}]
Read more : Which Goal Distinguishes Project Management And Portfolio Management
The Ka expression is:
If you are using the simpler version of the equilibrium . . .
[CH_3COOH rightleftharpoons CH_3COO^- + H^+ tag{8}]
. . . the Ka expression is:
The table shows some values of Ka for some simple acids:
acid Ka (mol dm-3) hydrofluoric acid 5.6 x 10-4 methanoic acid 1.6 x 10-4 ethanoic acid 1.7 x 10-5 hydrogen sulphide 8.9 x 10-8
These are all weak acids because the values for Ka are very small. They are listed in order of decreasing acid strength – the Ka values get smaller as you go down the table. However, if you aren’t very happy with numbers, that isn’t immediately obvious. Because the numbers are in two parts, there is too much to think about quickly! To avoid this, the numbers are often converted into a new, easier form, called pKa.
Source: https://t-tees.com
Category: WHICH
