How many moles of HCl must be added to 1.00 L of 0.72 M NH3 to make a buffer with a pH of 9.50?

Question

(Ka of NH4+ = 5.6e-10)

the answer is 0.26 mol but how did you get that?

Answer 1

@ pH = 9.50, the [H+] = 3.16e-10

as "X" amount of H+ was added , you lost "X" amount of NH3, & produced "X" amount of NH4+

=================

NH4+ -->  NH3  &  H+

X  --> --> 0.72-X & 3.16e-10

K = [NH3]  [H+] / [NH4+]

5.6e-10 = [0.72-X]  [3.16e-10] / [X]

5.6e-10 (X) = [0.72-X]  [3.16e-10]

5.6e-10 (X) = 2.28e-10 - 3.16e-10 (X)

8.76 e-10 (X) = 2.28e-10

X= 0.26 molar  H+, requires that you added 0.26 moles of H+ to the 1.00 litres

your answer is that 0.26 moles was added