Determine the ph of a buffer solution comprised of a 1.41 m hf and 0.623 m naf. the ka of hf = 7.20 x 10-4
2 answers:
<span>we are going to use H-H equation:
PH = Pka + log[conjugate base]/[weak acid]
when we have here the conjugate base is NaF
and the weak acid is HF.
so, we are going to use the Ka value to get the value of Pka.
when Pka = -logKa
= -log7.2 X 10^-4
= 3.14
then we will substitution on H-H equation:
</span>∴<span>PH = 3.14 + log (0.623)/(1.41) = 2.99</span>
Answer:
The pH of the buffer solution is 2.78.
Explanation:
Concentration of salt = 0.623 M
Concentration of acid = 1.41 M
Dissociation constant of acid = 
The pH of the buffer solution is given by Henderson-Hasselbalch equation:
![pH=pK_a+\log \frac{[salt]}{[acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D)
![pH=-\log[K_a]+\log \frac{salt}{acid}](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BK_a%5D%2B%5Clog%20%5Cfrac%7Bsalt%7D%7Bacid%7D)
![pH=-\log[7.20\times 10^{-4}]+\log\frac{0.623 M}{1.41 M}](https://tex.z-dn.net/?f=pH%3D-%5Clog%5B7.20%5Ctimes%2010%5E%7B-4%7D%5D%2B%5Clog%5Cfrac%7B0.623%20M%7D%7B1.41%20M%7D)
pH = 2.78
The pH of the buffer solution is 2.78.
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