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3 years ago

I need help on this, It’s due today. Please, anybody

Chemistry

1 answer:

Sholpan [36]3 years ago

4 0

Answer:

1 = 252g, 2 = 2mL, 3 = 1.5mL, 4 = 3g, 5 = 225g, 6 = 0.92g/mL, 7 = 0.75g/mL, 8 = 0.71g/mL, 9 = 1.9mL, 10= 1.11mL, 11 = 76.9g

Explanation:

This problem is testing how well you can move around the equation D = m/v where D = Density (g/mL), m= mass of sample (g), v = volume of sample (mL).

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

expeople1 [14]

Answer:

$2.72$ grams

Explanation:

Complete question is

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

(C6H5COOH)

and 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.

Solution

Given

pH of the buffer solution $= 4$

Concentration of C6H5COOH $= 0.02$ M

Volume of the buffer solution $= 1.50$ L

$K_a$ value for benzoic acid is $6.3 * 10^ {-5}$

Concentration of sodium benzoate

$pH = - log Ka + log \frac{C6H5COONa}{C6H5COOH}$

Substituting the given values we get

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

Number of moles in sodium benzoate

$= 0.0126 * 1.5 \\= 0.0189 Mol$

Mass of sodium benzoate

$0.0189 mol * 144.147 \frac{g}{mol} \\= 2.72 g$

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