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

Answer:

2.90

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:

$pH=pKa~+~Log\frac{[A^-]}{[HA]}$

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

$pH=~3.14~+~Log(\frac{[0.237~M]}{[0.413~M]})~=~2.90$

The pH value would be 2.90

I hope it helps!