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The acid dissociation constant of benzoic acid is 6.5 x 10^-5. Therefore, the pH of the benzoic acid solution prior to adding sodium benzoate is:

pH = -log[Ka]
pH = -log (6.5 x 10^-5)
pH = 4.19

The pH of the benzoic acid solution is 4.19 which is acidic, but a weak acid. 

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

Given data :

pH of buffer solution = 4

concentration of benzoic acid = 0.020 M

Volume of solution = 1.5 L

Ka for Benzoic acid = 6.3 * 10⁻⁵

Calculating the pH value of Benzoic acid solution prior adding sodium benzoate

lets represent the reaction

                               C₆H₅COOH(aq)  ⇄  H⁺ (aq) + C₆H₅COO⁻ (aq)

Eq concentration      0.0200 M - x              x       +      x

where :  Ka = [ H⁺ ] * [ C₆H₅COO⁻ ] / [ C₆H₅COOH ]

             6.3 * 10⁻⁵  = x * x / [ 0.0200 ]  ----- ( 1 )

Resolving equation 1

 x = 1.09 * 10⁻³ M

Hence ; [ H⁺ ] = 1.09 * 10⁻³ M

pH of benzoic acid  = -log [ H⁺ ]

     = - log [  1.09 * 10⁻³  ]

     = 2.96

Hence we can conclude that The pH of the benzoic acid solution prior to adding sodium benzoate is : 2.96

Learn more about Benzoic acid : https://brainly.com/question/3460480

The missing data related to your question is missing below is the complete question

You are asked to pre pare a pH = 4.00 buffer starting from 1.50 L of 0.0200 M solution of benzoic acid and any amount you need of sodium benzoate

What is the pH of the benzoic acid solution prior to adding sodium benzoate

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