Buffers questions help please

A student adds 50.0cm^3 of 0.250 mol dm^-3 butanoic acid to 50.0cm^3 of 0.0500 mol dm^-3 sodium hydroxide. A buffer solution forms.

i) Explain why a buffer solution forms

ii) Calculate the pH of the buffer solution.

The Ka of butanoic acid is 1.51 x a0^-5 mold dm^-3

Our teacher didnt teach us this method on making a buffer only the other one by adding a weak acid to its conjugate base

i) after the reaction between the acid and the base you are left with only the weak acid and its salt. So in this case you must do the stoichiometry calculation first to find the moles of acid remaining and the moles of salt formed.

You then use ka = [H+][A-]/[HA] as normal.

All the NaOh reacts so your left with

weak acid --> salt of weak acid+ H20 ?

... and also the excess acid (very important)

Buffer Concentrations
Conc HA = 1000/TOTAL VOLUME x 0.0125 = 1000/100 x 0.0125 = 0.125mol dm^-3
Conc A- = 1000/100 x 2.5 x 10^-3 = 0.025

The way I calculated it was:

c = n / v
c = concentration
n = number of moles
v = volume in cm3

[HA] = 0.0125 / (100/1000) = 0.125 mol dm^-3
[A-] = 2.5 x 10^-3 / (100/1000) = 0.025 mol dm^-3

I divided the 100cm3 by 1000 to get a value in dm3

Not sure if you still need help?

Forms a buffer as people have explained; HA + OH- --> A- + H2O

Calculation.

Moles of OH- = M x V / 1000 = 2.5 x 10^-3
Moles of A- thus equals 2.5 x 10^-3 (as shown in the initial equation)
Moles of HA = M x V / 1000 = 0.0125

Buffer Concentrations
Conc HA = 1000/TOTAL VOLUME x 0.0125 = 1000/100 x 0.0125 = 0.125mol dm^-3
Conc A- = 1000/100 x 2.5 x 10^-3 = 0.025

Rearrange Ka..

[H+] = Ka x [HA] / [A-]

Sub in.

[H+] = (1.51 x 10^-5) x (0.125) / (0.025)
=7.55 x 10^-5
-log10[H+]

ph = 4.12

-- i think--