[Yurij]

We titrate 100 ml of weak monoprotic acid using 27,63ml of 0.09381M NaOH. pH at eq. point is 10,99. Calculate pH after addition of 19,47ml of base to 100ml of acid.

pOH = 3.01

<OH^-> = 9.77 x 10^-4

We could also calculate initial molarity of the acid and moles of acid

n_a = n_b = 0,02763 L x 0,09381 mol x L^-1 = 2,6 x 10^-3 moles

now we could use HH equation since we're getting a buffer, but I don't know how to get Ka which is required for it

please help

[Borek]

pH at the equivalence point is a function of Ka and concentration.

[Yurij]

Is this right?

K_b = K_w/K_a = <OH->²/Cb

C_b = n_b/V_tot

K_a= (K_w x c_b)/<OH->² and from then we use HH Equation

[Borek]

Convoluted, but equivalent to what I had on mind.