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Suppose you want to prepare a buffer with a pH of 4.59 using formic acid. What ratio of [sodium formate]/[formic acid) do you need to make this buffer? Formic acid has a K, of 1.8x10 4 Enter your answer to three significant figures.

Sagot :

maacastrobr

Answer:

7.08

Explanation:

To solve this problem we'll use the Henderson-Hasselbach equation:

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

Where [tex]\frac{[A^-]}{[HA]}[/tex] is the ratio of [sodium formate]/[formic acid] and pka is equal to -log(Ka), meaning that:

  • pka = -log (1.8x10⁻⁴) = 3.74

We input the data:

  • 4.59 = 3.74 + log[tex]\frac{[A^-]}{[HA]}[/tex]

And solve for [tex]\frac{[A^-]}{[HA]}[/tex]:

  • 0.85 = log[tex]\frac{[A^-]}{[HA]}[/tex]
  • [tex]10^{(0.85)}[/tex]=[tex]\frac{[A^-]}{[HA]}[/tex]
  • [tex]\frac{[A^-]}{[HA]}[/tex] = 7.08
Eduard22sly

The ratio of [sodium formate] to [formic acid] needed to make the buffer is 7.08

How to determine the pKa

  • Equilibrium constant (Ka) = 1.8×10¯⁴
  • pKa =?

pKa = –Log Ka

pKa = –Log 1.8×10¯⁴

pKa = 3.74

How to determine the ratio of [sodium formate] to [formic acid]

  • pH = 4.59
  • pKa = 3.74
  • Ratio of [sodium formate] to [formic acid] =?

pH = pKa + Log[sodium formate]/[formic acid]

4.59 = 3.74 + Log[sodium formate]/[formic acid]

Collect like terms

4.59 – 3.74 = Log[sodium formate]/[formic acid]

0.85 = Log[sodium formate]/[formic acid]

Take the anti-log of 0.85

[sodium formate]/[formic acid] = anti-log 0.85

[sodium formate]/[formic acid] = 7.08

Learn more about pH of buffer:

https://brainly.com/question/3709867

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