You are instructed to create 300. mL of a 0.28 M Acetate buffer with a pH of 5.1

Question

You are instructed to create 300. mL of a 0.28 M Acetate buffer with a pH of 5.1. You have Acetic acid and the sodium salt NaC_2H_3O_2, available. (Enter all numerical answers to three significant figures.) Write the results of your calculations in your lab notebook for use during lab. HC_2H_2O_2(f) H_2O^+(aq) K_a1 - 1.7 times 10^-5 What is the molarity needed for the acid component of the buffer? M What is the molarity needed for the base component of the buffer? M How many moles of acid are needed for the buffer? mol How many moles of base are needed for the buffer? mol How many grams of acid are needed for the buffer? g How many grams of base are needed for the buffer? g

Explanation / Answer

we know that

pKa = -log Ka

pKa = -log 1.7 x 10-5

pKa = 4.76955

now

for buffers

pH = pKa + log [conjugate base / acid ]

in this case

pH = pKa + log [CH3COONa / CH3COOH]

so

5.1 = 4.76955 + log [CH3COONa / CH3COOH]

[CH3COONa / CH3COOH] = 2.14

[CH3COONa] = 2.14 [CH3COOH]

now

given

total acetate concentration = 0.28

so

[CH3COOH] + [CH3COONa] = 0.28

[CH3COOH] + 2.14 [CH3COOH] = 0.28

3.14 [CH3COOH] = 0.28

[CH3COOH] = 0.08917

now

[CH3COONa ] = 2.14 x 0.8917 = 0.19083 M

so

1) molarity of acid component CH3COOH is 0.08917 M

2) molarity of base component CH3COONa is 0.19083 M

now

we know that

moles = molarity x volume (ml) / 1000

so

3) moles of acid , CH3COOH = 0.08917 x 300 / 1000 = 0.026751 mol

4) moles of base , CH3COONa = 0.19083 x 300 / 1000 = 0.057249 mol

now

we know that

mass = moles x molar mass

so

5) moles of acid , CH3COOH = 0.026751 x 60 = 1.60506 g

6) moles of base , CH3COONa =0.057249 x 82 = 4.694418 g

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