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

Question

You 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 CH_3COOH _____ Grams NaOOCCH_3 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. 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 each occurred.

Explanation / Answer

3.

Molarity = Moles / Liter

Molarity in question 1 are

[CH3COOH] = 0.150 M

[CH3COONa] = 0.250 M

20 mL of 0.150 M CH3COOH

So, moles of CH3COOH = 0.02 L x 0.150 M = 0.003 moles = 3 mmol

20 mL of 0.250 M CH3COONa

So, moles of CH3COONa = 0.02 L x 0.250 M = 0.005 moles = 5 mmol

Now

1.0 mL of 1.0 M HCl

So, moles of HCl = 0.001 L x 1.0 M = 0.001 moles

When 0.001 moles of HCl is added, it will react with 0.001 moles of CH3COONa to produce 0.001 moles of CH3COOH. So, moles of CH3COONa will decrease by 0.001 moles and moles of CH3COOH will increase by 0.001 moles.

CH3COONa    +    HCl    =    CH3COONa    +    NaCl

Moles after addition of HCl are

moles of CH3COOH = 0.003 moles + 0.001 mole = 0.004 moles = 4 mmol

moles of CH3COONa = 0.005 moles – 0.001 mole = 0.004 moles = 4 mmol

You asked for question number 3 only.

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