Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.

QUESTION

# 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.

LEARN MORE EFFECTIVELY AND GET BETTER GRADES!