Question

The following chemical species are mixed:

[ H+] = 0.25 M

[NH3] = 0.25 M

What kind is formed in the balance?

And what is the pH of the solution?

Answer 1

Solution :-

When the NH3 is mixed with H^+ then it forms conjugate acid NH4^+ by the following equation

NH3 + H^+ ---- > NH4^+

so it produces the conjugate acid (NH4^+) of the weak base NH3

Since the concentrations are same therefore their volume will also be same

Lets assume 1 L of each species are mixed then at the total volume new molarity of the NH4^+ will be calculated as

[NH4^+] = 0.25 M* 1 L / 2.0 L = 0.125 M

Now lets set up the ICE table for the NH4^+

NH4^+ + H2O ----- > NH3 + H3O^+

0.125 M                         0            0

-x                                     +x          +x

0.125-x                            x            x

Ka of NH4^+ = 5.6*10^-10

Ka=[H3O^+][NH3]/[NH4^+]

5.6*10^-10 = [x][x]/[0.125-x]

Since the Ka is very small therefore we can neglect the x from denominator

5.6*10^-10 = [x][x]/[0.125]

5.6*10^-10 * 0.125 =x^2

7.0*10^-11 = x^2

Taking square root on both sides we get

8.37*10^-4 = x =[H3O^+]

Now lets calculate the pH

pH= -log [H3O^+]

pH= -log [8.37*10^-6]

pH= 5.08

Therefore pH after the mixing will be 5.08