Question

A buffer solution is 0.413 M in HF and 0.237 M in KF. If Ka for HF is 7.2×10-4, what is the pH of this buffer solution?

Answer 1

2.90

Step-by-step explanation:

Any buffer system can be described with the reaction:

`HA~->~H^+~+~A^-`

Where
`HA` is the acid and
`A^-` is the base. Additionally, the calculation of the pH of any buffer system can be made with the Henderson-Hasselbach equation:

`pH=pKa~+~Log([A^-])/([HA])`

With all this in mind, we can write the reaction for our buffer system:

`HF~->~H^+~+~F^-`

In this case, the acid is
`HF` with a concentration of 0.413 M and the base is
`F^-` with a concentration of 0.237 M. We can calculate the pKa value if we do the "-Log Ka", so:

`pKa~=~-Log(7.2X10^-^4)=~3.14`

Now, we can plug the values into the Henderson-Hasselbach

`pH=~3.14~+~Log(([0.237~M])/([0.413~M]))~=~2.90`

The pH value would be 2.90

I hope it helps!