1You have been asked to prepare 100.0 mL of a buffer solution that is 0.150 M in

Question

1You have been asked to prepare 100.0 mL of a buffer solution that is 0.150 M in acetic acid and 0.250 M in sodium acetate. Calculate the grams of each component that you would weigh out. Show your work below. Grams CHsCOOH 2. To 20.0 mL of the buffer in question 1, you add 1.0 mL of 1.00 M NaOH. Write the chemical equation for the neutralization reaction. Calculate the mmol of each reactant and product before the reaction occurs and after the reaction has occurred Grams NaOOCCH 3. To 20.0 mL of the buffer in question 1, you add 1.0 mL of 1.00 M HCL Write the chemical equation for the neutralization reaction. Calculate the mmol of each reactant and product before the reaction occurs and after the reaction has occurred

Explanation / Answer

1) Total volume of buffer = 100mL

the concentration of acetic acid = 0.150M

We know molarity = number of moles /Volume(L)

= (mass / molar mass) /VL

0.15 0 = (mass?60g/mol) / 0.1

thus mass of acetic acid = 0.15x60 x0.1 g

= 0.9 g

Similarly of mass of sodium acetate in 100mL solution of 0.250M

0.250M = (mass/82g/mol) /0.1

and mass of sodium acetate to be taken = 0.25 x82x 0.1 g

= 2.05 g

2) To 20mL of this buffer 1.0 ml of 1.0M NaOh is added

CH3COOH + OH- ----------------> CH3COO- + H2O

20x0.15 0 20x0.25 - initial mmoles

=3 = 5

- 1 - change

2 0 6 after reaction

3)

Now 1.0mL of 1.0M HCl is added to 20 mL of above buffer

CH3COO- + H+ --------------------> CH3COOH

20x0.25= 5 0 20x0.15=3 initial mmoles

- 1x1.0 - change

4 0 4 after reaction

  

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