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What is the pH of a buffer prepared by adding 0.809 mol of the weak acid HA to 0.406 mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66×10−7.

Respuesta :

Answer:

5.94

Explanation:

Given:-

[tex]K_{a}=5.66\times 10^{-7}[/tex]

[tex]pK_{a}=-\log\ K_{a}=-\log\ (5.66\times 10^{-7})=6.24[/tex]

[salt] = 0.406 mol

[acid] = 0.809 mol

Considering the Henderson- Hasselbalch equation for the calculation of the pH of the acidic buffer solution as:

[tex] pH=pK_a+log\frac{[salt]}{acid]} [/tex]

So,  

[tex] pH=6.24+\log (\frac{0.406}{0.809})=5.94 [/tex]

Eduard22sly

The pH of the buffer solution is 5.95

How to determine the pKa

We'll begin by calculating the pKa of the solution. This can be obtained as follow:

  • Equilibrium constant (Ka) = 5.66×10¯⁷
  • pKa =?

pKa = –Log Ka

pKa = –Log 5.66×10¯⁷

pKa = 6.25

How to determine the pH

  • pKa = 6.25
  • Mole of acid = 0.809 mole
  • Molarity of acid [HA] = 0.809 / 2 = 0.4045 M
  • Mole of salt = 0.406 mole
  • Molarity of salt [NaA] = 0.406 / 2 = 0.203 M
  • pH =?

pH = pKa + Log[salt]/[acid]

pH = 6.25 + Log[0.203]/[0.4045]

pH = 5.95

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