Question

We're asked to find the pH of 50mL of .1 M of a weak acid (labeled HX), before we add any .1 M NaOH. The Ka is 1.5e-5. The pH I found was 4.00 (feel free to fact check me). Here's the part I need help with:

-We are also asked to find the pH of the solution at the "endpoint". Not sure what this means. Do they mean equilibrium? If so, how would I calculate this? Thanks.

Answer 1

a given c = (Hx] = 0.1m V = somi Ka = 105x105 it is a weak acid pH = Ź (pka lege] = {(4.823-legen) (*pe-bay ra = 2.911 { = -log('5 x 107 1 = 4.823 here end point means acid its Completely weet valised and we will be - at having only salt HX + Nach •1X50 clxu end point Nax+ H2O cat end point both had lame •1x50 = cхk [= 50ml 100 milimees) Tatal volume = 50+50=100ml c'= (Max] = 5 = 0.05m *: pH of WASB (kibax acid strangbale) sait pH = 7+ 1 2 (pka + lage) i 7+4 ( 4.823 +log(0.05)) 6 . pH= 876