You are asked to prepare a pH=4.00 buffer starting from 1.50 L of 0.0200 M solution of benzoic acid (C6H5COOH) and an excess of sodium benzoate (C6H5COONa). You may want to reference (Pages 721 - 728) Section 17.2 while completing this problem. Part A What is the pH of the benzoic acid solution prior to adding sodium benzoate?

Respuesta :

Answer:

pH = 2.96

Explanation:

  • C6H5COOH  ↔ C6H5COO-  + H3O+

∴ Ka = 6.5 E-5 = [H3O+][C6H5COO-] / [C6H5COOH]

  • C6H5COONa → C6H5COO-  +  Na+

prior to adding C6H5COONa:

C C6H5COONa = 0 M = [Na+]

∴ mass balance:

C C6H5COOH = [C6H5COOH] + [C6H5COO-] = 0.0200 M

⇒ [C6H5COOH] = 0.0200 - [C6H5COO-]

charge balance:

⇒ [H3O+] = [C6H5COO-]

⇒ [C6H5COOH] = 0.0200 - [H3O+]

⇒ Ka = [H3O+]² / (0.0200 - [H3O+]) = 6.5 E-5

⇒ [H3O+]² + 6.5 E-5[H3O+] - 1.3 E-6 = 0

⇒ [H3O+] = 1.108 E-3 M

∴ pH = - Log [H3O+]

⇒ pH = 2.9554 = 2.96

The pH of the benzoic acid solution prior to adding sodium benzoate is 2.96.

What is pH?

pH means potential hydrogen. The pH tells how acidic or basic a solution is.

Given,

For benzoic acid

[tex]\bold{C_6H_5COOH \rightarrow C_6H_5COO- + H_3O+}[/tex]

The ka value will be

[tex]\bold{6.5 E-5 = [H_3O+][C_6H_5COO-] / [C_6H_5COOH]}[/tex]

   

For sodium benzoate

[tex]\bold{C_6H_5COONa \rightarrow C_6H_5COO- + Na+}[/tex]

The Mass balance

[tex]\bold{ C_6H_5COOH [C_6H_5COOH] + [C_6H_5COO-] =0.0200\; M}\\\\\\\bold{[C_6H_5COOH] = 0.0200 - [C_6H_5COO-]}[/tex]

The charge balance

[tex][H_3O+] = [C_6H_5COO-] =[C_6H_5COOH] = 0.0200 - [H_3O+][/tex]

[tex]\bold{Ka =\dfrac{[H_3O+]²}{ (0.0200 - [H_3O+]) } = 6.5 E-5}[/tex]

[tex][H_3O+]^2+ 6.5 E-5[H_3O+] - 1.3 E-6 = 0\\[/tex]

[tex][H_3O+] = 1.108 E-3 M\\\\pH = - Log [H_3O+]\\\\pH = 2.9554 = 2.96[/tex]

Thus, the pH is 2.96.

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