Because HNO3 is a strong acid, [H+] = 4.2 x 10-3 M.  The pH of a solution is equal to the negative log of the concentration. In this case, pH = –log(4.2 x 10-3). Using the approximation, we can solve –log(4.2 x 10-3) ≈ 3 - .42 ≈ 2.5, which is closest to answer b).

An important thing to note in this problem is that, when an acid is in a solution containing equal quantities of the acid and conjugate base, the pH is equal to the pK_a.

Therefore, we have pH = 3.62 = pK_a. However, we are trying to solve for the K_a, not the pK_a (recall also that pK_a = –log(K_a) ), and we will need to use the approximation we learned before but in reverse.

pH = n – 0.m ≈ –log(m x 10^-n)

We can rewrite the pH of 3.62 as 4 – 0.38, putting it in the form n – 0.m shown above.  This gives us

pH = 4 – 0.38 ≈ –log(m x 10^-n)

Now we can substitute in m and n to approximate K_a on the right hand side.

pH = 4 – 0.38 ≈ –log(3.8 x 10^-4)

So the approximate value of the K_a is 3.8 x 10^-4 M, which is closest to answer c).