What is the pH of a 200.00 mL solution containing 21.46 g of benzoic acid and 37

Question

What is the pH of a 200.00 mL solution containing 21.46 g of benzoic acid and 37.68 g of sodium benzoate.( Ka = 6.5 x 10-5) Initially?! After 30.0 ml of 5.00 HCL has been added?! Has the buffer exceeded its capacity?! What is the pH of a 200.00 mL solution containing 21.46 g of benzoic acid and 37.68 g of sodium benzoate.( Ka = 6.5 x 10-5) Initially?! After 30.0 ml of 5.00 HCL has been added?! Has the buffer exceeded its capacity?! What is the pH of a 200.00 mL solution containing 21.46 g of benzoic acid and 37.68 g of sodium benzoate.( Ka = 6.5 x 10-5) Initially?! After 30.0 ml of 5.00 HCL has been added?! Has the buffer exceeded its capacity?!

Explanation / Answer

CALCULATION OF pH OF THE BUFFER SOLUTION BEFORE ADDING HCl

The pH of a buffer solution is given by the Henderson Hasselbalch equation

pH = pKa + log [salt]/[acid]

pKa = - log10 Ka = - log10 6.5x10-5

       = 4.19

[salt] = concentration of the salt, sodium benzoate in mole per liter (M)

CALCULATION OF [salt]

Molecular mass of sodium benzoate, C6H5COONa = 7 x12 + 5x1 + 2x16+ 23 = 144

Number of moles of C6H5COONa = Mass of C6H5COONa/ Molecular mass of C6H5COONa

                                                  = 37.68/144 = 0.2617

0.2617 moles of C6H5COONa is present in 200 mL solution

Number of moles of C6H5COONa present in 200 mL solution = 0.2617

Number of moles of C6H5COONa present in 1000 mL solution = 0.2617x1000/200 = 1.308

[acid] = concentration of the acid benzoic acid, C6H5COOH in mole per liter (M)

CALCULATION OF [acid]

Molecular mass of benzoic acid, C6H5COOH = 7 x12 + 6 x1 + 2x16 = 122

Number of moles of C6H5COOH = Mass of C6H5COOH/ Molecular mass of C6H5COOH

                                                  = 21.46/122 = 0.1759

Number of moles of C6H5COOH present in 200 mL solution = 0.1759

Number of moles of C6H5COONa present in 1000 mL solution = 0.1759 x1000/200 = 0.8795

The pH of a buffer solution

pH = pKa + log [salt]/[acid]  

     = 4.19 + log 1.308/0.8795

     = 4.36

CALCULATION OF pH OF THE BUFFER SOLUTION AFTER ADDING HCl

When a strong acid like HCl is added to an acid buffer,it will react with the salt to form undissociated weak acid.

C6H5COO- + H+ = C6H5COOH

**** As a result the concentration of the salt will decrease and that of the acid will increase .

The amount of acid added = 30.0 mL of 5 M HCl = 150 millimole = 0.150 mole

Assuming HCl to be completely dissociated, the amount of H+ = 0.150 mole

After adding HCl, the concentration of the salt will decrease and that of the acid will increase .

[ C6H5COO-] = [salt] = 1.308 mole - 0.150 mole = 1.158 mole/L

[ C6H5COOH] =[acid] = 0.8795 mole + 0.150 mole = 1.029 mole/L

pH = pKa + log [salt]/[acid]

     = 4.19 + log 1.158/1.029 = 4.24

Here it is ssumed that the change in volume from the addtion of HCl is negligible

pH OF THE BUFFER SOLUTION BEFORE ADDING HCl = 4.36

pH OF THE BUFFER SOLUTION AFTER ADDING HCl = 4.24

CHANGE OF pH = 4.36 - 4.24 = 0.12

There is only a slight change in the pH.

THE BUFFER HAS NOT EXCEEDED ITS CAPACITY

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