Question

What is the pH of a solution that results from mixing together equal volumes of a 0.050 M solution of ammonia and a 0.025 M solution of hydrochloric acid? (pKa of NH4+ = 9.25)

Answer 1

Let, 1000ml 0.025M HCl and 1000 ml 0.050M NH₃ are made to react.

Given, pKa = 9.25

Now, 0.025 mol HCl will produce 0.025 mol NH4Cl in the final solution but the volume of the final mixture is now (1000+1000) = 2000 ml.

So, in the final solution contains, 0.025 mol NH3 and 0.025 mol NH4+, i.e, a buffer solution , so it'll follow Henderson Equation, which is the following,

pOH= pKb + log{[salt]/[base]}

Now, pH + pOH = 14 and pKa + pKb= 14, at 298K

So we can write,

(14 - pH) = (14 - pKa) + log{[salt]/[base]}

or, pH = pKa + log{[base]/[salt]}.....(i)

Now, we know, V₁S₁ = V₂S₂

where, V = volume and S = strength or concentration of the corresponding solution and the subscripts 1,2 are just to denote the different volumes and concentrations.

So, concentration of salt, [salt]=[(0.025*1000)/2000]=0.0125M

concentration of base, [base]=[(0.025*1000)/2000]= 0.0125M.

Therefore, putting the values in equation (i) we get,

pH = pKa + log{0.0125/0.0125}

or, pH = 9.25 + log{ 1 }

So, pH = 9.25

So, the final pH of the solution would be 9.25.

Note: for the sake of simplicity we take the volume of ammonia and HCl as 1000 ml. One can take any arbitrary volume ; the result will exactly be the same as above.