How many grams of base are needed for the buffer? You are instructed to create 1

Question

How many grams of base are needed for the buffer? You are instructed to create 170. mL of a 0.06 M Acetate buffer with a pH of 4.1. 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_3O_2(s) + H_2O(l) H_3O^+(aq) + C_2H_3O_2^-(aq) K_a1 = 1.7 times 10^-5 What is the molarity needed for the acid component of the buffer? What is he molarity needed for the base component of the buffer? How many moles of acid are needed for the buffer? How many moles of base are needed for the buffer? How many grams of acid are needed for the buffer? How many grams of base a needed for the buffer?

Explanation / Answer

HC2H3O2 (s) + H2O (l) <======> H3O^+(aq) + C2H3O2^ (aq)

To create170.0 mL of a 0.06 M acetate buffer, total moles of buffer = 170 ml x 0.06M = 10.2 m moles

PH of buffer = 4.1

Ka1 = 1.710^5

PKa = -log (Ka)

PKa = - log (1.710^5) = 4.8

We can calculate (base)/(acid) by the Henderson-Hasselbalch equation

PH = PKa + log (base)/( acid)

4.1 = 4.77+ log (base)/(acid)

log (base)/(acid) = 4.77 - 4.1 = 0.67

(base)/ (acid) = 10^ 0.67

(base)/ (acid) = 4.677

(Acid + base) = 10.2 m moles

Moles of acid = 1.796 m moles of acid = 0.0018 moles

Moles of base = 8.385 m moles 0.0084 moles

i)The molarity needed for the acid component of the buffer = moles of acid/volume of buffer in L

The molarity of acid = 0.0018 moles/0.170 L = 0.0106M

ii) The molarity needed for the base component of the buffer = moles of base/volume of buffer in L

The molarity of base = 0.0084 moles/0.170 L = 0.0494 M

iii) Moles of acid are needed for the buffer -  

(Base)/ (acid) = 4.677

(Acid + base) = 10.2 m moles

1 a + 4.677a = 10.2

5.677a = 10.2 a = 10.2/5.677 = 1.796 m moles

Now convert m. mole to moles dividing by 1000 = 0.001776 moles

Moles of acid are needed for the buffer    = 0.00180 moles

iv) Mole of base = total moles of buffer – moles of acid

Moles of base = 10.2 m moles – 1.8 m moles = 8.4 m moles

Now convert m. mole to moles dividing by 1000 = 0.0084 moles

Moles of base are needed for the buffer = 0.00840 moles

v) Grams of acid are needed for the buffer -

Molar mass of acetic acid = 60.05 g/mol

Grams of acid = Moles of acid x Molar mass of acetic acid

0.0018 moles x 60.05 g/mol = 0.108 g

vi) Grams of base are needed for the buffer

Molar mass of NaC2H3O2 is 82.0338 g/mol.

Grams of base = Moles of base x Molar mass of base

Grams of base = 0.0084 moles x 82.0338 g/mol. = 0.689 g

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