Is this a Buffer? 25 ml of 0.60 M HClO4 + 45 ml 0.60 M KF. Please show how you determined its a buffer and mention the concentration of weak base and weak acid. PH= Pka + log(weak base/weak acid). (buffer equation)

Yes, this is a buffer with = 3.04.

##"KF"## is the salt of a strong base and the weak acid ##"HF"##.

The ##"F"^-## ion is basic and will react completely with a strong acid like ##"HClO"_4##.

The ##color(red)("molecular equation")## is

##"KF(aq)" + "HClO"_4"(aq)" → "HF(aq)" + "KClO"_4"(aq)"##

The ##color(red)("ionic equation"## is

##"K"^+"(aq)" + "F"^(-)"(aq)" + "H"_3"O"^+"(aq)" + "ClO"_4^(-)"(aq)" → "HF(aq)" + "H"_2"O(l)" + "K"^+"(aq)" + "ClO"_4^(-)"(aq)"##

The ##color(red)("net ionic equation"## is

##"F"^(-)"(aq)" + "H"_3"O"^+"(aq)" → "HF(aq)" + "H"_2"O(l)"##

The first problem is to figure out how much ##"F"^-## reacts and how much is unreacted.

##"Moles of F"^(-) = 0.045 cancel("L") × "0.60 mol"/(1 cancel("L")) = "0.027 mol"##

##"Moles of H"_3"O"^+ = 0.025 cancel("L") × "0.60 mol"/(1 cancel("L")) = "0.0015 mol"##

We have a buffer, because we have a solution of a weak acid ##"HF"## and its conjugate base ##"F"^-##.

To calculate the pH, we use the Henderson-Hasselbalch Equation.

##"HF" + "H"_2"O" → "H"_3"O"⁺ + "F"^-##; ##"p"K_"a" = 3.14##

##"pH" = "p"K_"a" + log(("[F"^(-)"]")/"[HF]") = 3.14 + log((0.012 cancel("mol"))/(0.015 cancel("mol"))) = 3.14 – 0.097 = 3.04##

Note that the ratio of the concentrations is the same as the ratio of the moles, because both species are in the same solution.