Question

Find the activity coefficient for h+ in a solution of 0.009 (I) Al(NO3)3 and 0.004 M HCl. (Both dissolve completely and you may assume that the [OH-] is negligible. Use the nearest column for u on Table 7.1). What is the pH of the solution

Answer 1

Step 1: Find ionic strength, I

I = 1/2(cz^2)

   = ½ [ 0.009 x 1 (+3)² + 0.009 x 3 x(-1)² + 0.004 x 1 (+1)² + 0.004 x 1 (-1)²]

   = ½ [0.081 + 0.027 + 0.004 + 0.004]

= 0.058 M

Step : the activity coefficient equation

where a* is the size of H⁺ ion = 900 pm = 9 x 10^-10 m

(a* value from literature which I found, in question it is not given)

log () = - [0.5 x (1)² x (0.058)^1/2] / [ 1 + ( 9 x 10^-10 x (0.058)^1/2 / 305)]

=  - [0.5 x 0.24] / [ 1 + ( 7.1 x 10^-13)]

= - 0.12 / 1 ; neglecting ( 7.1 x 10^-13) term since its too small.

= - 0.12

So, = 10^-0.12 = 0.76

step 3: You can use this activity coefficient and [H+] to find pH...

pH = - [log(0.004 M) +log(0.76)]

= - [ -2.4 - 0.12]

= 2.52