A buffer is prepared by adding 0.40 moles of formic acid (HCOOH) and 0.60 moles of sodium formate (HCOONa) to enough water to make 1.00 L of solution.a. Calculate the pH of this buffer.b. How many moles of HCOONa must be added to 1.0 L of a 0.40M HCOOH solution to prepare a buffer with pH 4.24?c. Calculate the change in pH for 100mL of the buffers from part a) and b) upon the addition of 5.00 mL of 1.00M HCl.

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Asked Dec 13, 2019
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A buffer is prepared by adding 0.40 moles of formic acid (HCOOH) and 0.60 moles of sodium formate (HCOONa) to enough water to make 1.00 L of solution.

a. Calculate the pH of this buffer.

b. How many moles of HCOONa must be added to 1.0 L of a 0.40M HCOOH solution to prepare a buffer with pH 4.24?

c. Calculate the change in pH for 100mL of the buffers from part a) and b) upon the addition of 5.00 mL of 1.00M HCl. 

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Expert Answer

Step 1

A solution which resists the change in pH when a small amount of base or acid is added to it is known as a buffer solution. The behaviour of a buffer is described by the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation for a mixture of weak acids and its salts with a strong base is shown below:

 

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[Salt] weak acid] pH=pKa +log;

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Step 2

The standard Ka of HCOOH is 1.77×10-4. The pKa is the negative logarithm of the dissociation constant of the weak acid which is expressed by equation (1).

Substitute, 1.77×10-4 in equation (1) as shown in equation (2).

 

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pKa =-log (Ka)... (1) pKa =-log 1.77×10–4 . (2) = 3.75

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Step 3

(a)

Calculate the resultant pH of the solution using the Henderson-Hasselbalch equation.The Henderson-Hasselbalch equation for the given reaction is expressed...

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[HCOONA НСООН pH=pKa + log (3) [0.60 M] [0.40 M] pH=3.75+log .. (4) = 3.93

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