calculate the pH at the stoichiometric point of the titration of 25.00 ml of 0.020M NH3 with 0.015 HCL (AQ)?

Question

for NH4^+
use ka = 5.6x10^10

Answer 1

The initial reaction is NH3 + H+ -----> NH4+, so it's 1:1. If you have 25 ml of 0.020 M, which is 0.5 mmol, so you would need 0.5 mmol of HCl. From a 0.015 M solution, that is 33.33 ml, so your total volume is 58.33 ml.

Thus your concentration of NH4+ to determine pH at the equivalence point is 0.5 mmol/58.33 ml or 0.00857 M.

Ka = [NH3][H+]/[NH4+]
5.6*10^-10 = x^2/0.00857-x x<<<0.0857, so drops out
4.8*10^-12 = x^2
x = [H+] = 2.19 * 10^-6; pH = 5.66

Answer 2

NH3 is a weak base.
when we titr a sol. with another, an indicator is used to show the end point of titration. we add enough amount of HCl till inicator color changes and the final pH depends on the indicator. at that point all the NH3 is titred regardless of it's equilibrium constant. the aim for titration is finding the volume of titrant needed 25*0.02=x(ml)*.015 --> x=33.3ml