If an acid has a ka =1.6 x 10-10 what is the acidity of the solution

Imagine the hypothetical acid HA whose equilibrium is

           HnAm (aq) ⇄ nH+    +   m A-

And the Ka is, Ka = [H+]^n * [A-]^m / [HnAm]

Aciditiy is a measure of the [H+] in equilibbrium.

The higher [H+] the higher the acidity. 

And from the equation

- the higher [H+] the higher Ka,
- the lower [H+] the lower Ka
- the higher Ka the higher [H+].
- the lower Ka the lower [H+]

Then, you conclude that for a low Ka (which 1.6 * 10^-10 is) the acidity is low.

That means that the corresponding acid has a low acidity (it is a weak acid).      

You can calculte the acidity as the pKa

pKa  = log { 1 / Ka} = log { 1 / ( 1.6*10^-10) } =  9.8

the acidity is low with a pKa of 9.8.

Hi; /c

Neither model is able to show the reactivity of potassium.