How many grams of sodium benzoate should be added to prepare the buffer? Neglect the small volume change that occurs when the sodium benzoate is added.

Solution

Given

pH of the buffer solution [tex]= 4[/tex]

Concentration of C6H5COOH [tex]= 0.02[/tex] M

Volume of the buffer solution [tex]= 1.50[/tex] L

[tex]K_a[/tex] value for benzoic acid is [tex]6.3 * 10^ {-5}[/tex]

Concentration of sodium benzoate

[tex]pH = - log Ka + log \frac{C6H5COONa}{C6H5COOH}[/tex]

Substituting the given values we get

[tex]log \frac{C6H5COONa}{C6H5COOH} = 4 + log (6.3 * 10^ {-5})\\log \frac{C6H5COONa}{C6H5COOH} = -0.20\\\frac{C6H5COONa}{C6H5COOH} = 10^{-0.2}\\{C6H5COONa} = 10^{-0.2} * 0.02\\{C6H5COONa} = 0.63 * 0.02\\{C6H5COONa} = 0.0126 M[/tex]

Number of moles in sodium benzoate

[tex]= 0.0126 * 1.5 \\= 0.0189 Mol[/tex]

Mass of sodium benzoate

[tex]0.0189 mol * 144.147 \frac{g}{mol} \\= 2.72 g[/tex]

batolisis batolisis · Answer 2 · 2022-04-05T13:35:50+02:00

The grams of sodium benzoate that should be added to prepare the buffer is : 2.72 grams

Given data :

pH of solution = 4

Concentration of C₆H₅COOH = 0.02 M

Volume of buffer solution = 1.5 L

Ka value = 6.3 * 10⁻⁵

Determine the mass of sodium benzoate that should be added to prepare the buffer

First step : determine the concentration of sodium benzoate

Apply the equation below

pH = -log Ka + log [tex]\frac{C_{6}H_{5}COONa }{C_{6}H_{5}COOH }[/tex] ---- ( 1 )

from equation ( 1 )

4 = - log ( 6.3 * 10⁻⁵ ) + log [tex]\frac{C_{6}H_{5}COONa }{0.02 }[/tex]

therefore

Concentration of [tex]C_{6}H_{5}COONa[/tex] = 0.63 * 0.02

= 0.0126M

Next step : Determine the number of moles

number of moles of sodium benzoate = 0.0126 * 1.5

= 0.0189 Mol

Final step : determine the mass of the sodium benzoate

mass of sodium benzoate = 0.0189 * 144.147

= 2.72 grams

Hence we can conclude that The grams of sodium benzoate that should be added to prepare the buffer is : 2.72 grams

Learn more about sodium benzoate : https://brainly.com/question/6471624

Attached below is the complete question

You are asked to prepare a pH = 4.00 buffer starting from 1.50 L of 0.0200 M solution of benzoic acid

(C6H5COOH)

add any amount you need of sodium benzoate

(C6H5COONa)

How many grams of sodium benzoate should be added to prepare the buffer? Neglect the small volume change that occurs when the sodium benzoate is added.