How to find the pH when adding unknown amounts of weak acid into known amounts of strong base??

It could be done if the concentration of the acid is known from the start. The #K_a# does not say anything about the starting concentration of the acid, and thus, we do not know how many mols react...

This is why we react strong base INTO weak acid, not the other way around.

Suppose it was #"0.1 M"# weak acid. You start off with #"0.1 M NaOH"#, for which

#"pOH" = -log["OH"^(-)] = 1#

and

#color(black)("pH") = 14 - 1 = color(blue)(13)#

Then you react weak acid into strong base? Well...

#"0.1 mol"/"L" xx "0.005 L" = "0.0005 mols butanoic acid"#

#"0.1 mol"/"L" xx "0.025 L" = "0.0025 mols OH"^(-)#

That means only #"0.0005 mols"# of #"OH"^(-)# reacts, so #"0.0020 mols"# remains, in a total of #"30 mL"#. That generates #"0.0005 mols A"^(-)#, whereas the #"0.0005 mols"# of #"HA"# is gone already.

The #"A"^(-)# could back-associate I suppose...

#"A"^(-) + "H"_2"O"(l) rightleftharpoons "HA"(aq) + "OH"^(-)(aq)#

#K_b = K_w/K_a = 10^(-14)/(1.5 xx 10^(-5)) = 6.67 xx 10^(-10)#

#= (["HA"]["OH"^(-)])/(["A"^(-)]) = x^2/("0.0005 mols A"^(-)/"0.030 L" - x)#

#~~ x^2/("0.0167 M")#

Hence,

#x = 3.34 xx 10^(-6) "M OH"^(-)# from #"A"^(-)#

But since that is on the order of #10^(-7) "M"#, it would be swallowed up by the autoionization of water. Thus,

#("0.0020 mols OH"^(-) + "negligible OH"^(-) "from " "A"^(-))/"0.030 L" = "0.067 M"#

And so, the #"pOH"# is

#"pOH" = -log["OH"^(-)] = 1.18#

and

#color(blue)("pH") = 14 - 1.18 = color(blue)(12.82)#