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Consider a buffer made by adding 132.8 g of NaC₇H₅O₂ to 300.0 mL of 1.17 M HC₇H₅O₂ (Ka = 6.3 x 10⁻⁵) What is the pH of this buffer?

Answer :

maacastrobr

Answer:

pH = 1.62

Explanation:

We can solve this problem using Henderson-Hasselbach's equation:

  • pH = pKa + log[tex]\frac{[A^-]}{[HA]}[/tex]

Where

  • pKa = -log(Ka) = 4.20
  • [A⁻] = [NaC₇H₅O₂]
  • [HA] = [HC₇H₅O₂]

Now we convert 132.8 g of NaC₇H₅O₂ into moles, using its molar mass:

  • 132.8 ÷ 144.11 g/mol = 0.921 moles

Then we calculate [NaC₇H₅O₂]:

  • 0.921 moles / 300.0 mL = 0.003 M

Now we can proceed to calculate the pH of the solution:

  • pH = 4.20 + log[tex]\frac{0.003}{1.17}[/tex]
  • pH = 1.62

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