What is the pH of a 0.1 M acid solution?

1 Answer
Jun 9, 2016

For a strong acid, #pH# #=# #-log_10[H_3O^+]# #=# #-log_10[0.1]=-log_10(10^-1)# #=# #-(-1)# #=# #1#

Explanation:

We have assumed a strong acid, for which the equilibrium reaction, as shown below, lies strongly to the right.

#HX(aq)+H_2O(l)rightleftharpoonsX^(-) + H_3O^(+)#

Why do we use #pH#? Well it is a holdover from the pre-electronic calculator days, which some of us can remember (not me of course, I am not a day over 29!).

When I write #log_ab=c#, I explicitly say that #a^c=b#. Common logarithmic bases are #10# and #e# (I think you cover this in A level mathematics these days). #pH# and #pK_a# scales use logarithms to the base #10#.

Anyway given what I have said #log_(10)100=2#, #log_(10)1000=3#, and #log_(10)1=0#, and #log_(10)0.1=log_(10)10^-1=-1#. If you can grasp this, you will find the #pH# concept (literally #"pouvoir hydrogene, power of hydrogen"#) fairly straightforward.

With weaker acids, say acetic acid, #H_3C-C(=O)OH#, the acid-base equilibrium lies to the left. The equilibrium constant, #K_a#, can be measured and often reported as #pK_a# #=# #-log_10K_a#.