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kap26 [50]
3 years ago
10

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?

Chemistry
1 answer:
muminat3 years ago
6 0

Answer:

2.90

Explanation:

Any <u>buffer system</u> can be described with the reaction:

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

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

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 <u>pKa value</u> 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

<u>The pH value would be 2.90</u>

I hope it helps!

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