Question

4. Buffer capacity refers to the amount of acid or base a buffer can absorb without a significant pH change. It is governed by the concentrations of the conjugate acid and base components of the buffer. A 0.5 M buffer can "absorb five times as much acid or base as a 0.1 M buffer for a iven p calculate the new pH after a particular quantity of acid or base is ad You are given 60 mL of 0.50 M phosphate buffer, pH 6.83, to test. The starting composition ration and ded. of the buffer, both i phosphate species, is: Concentration of HPO:0.304 M Concentration of H,PO: 0.196 M a. You add 1.7 mL of 1.00 M HCl to the buffer. Calculate the molar quantity of H,O* added as Molar quantity of HPO: 18.2 mmol Molar quantity of H,Po,: 11.8 mmol HCI, and the final molar quantity of HP042. and H2POį at equilibrium. what is the new HP0MH,POİ ratio, and the new pH of the solution? The pK, of HPoi s 6.64. molar ratio rather than the concentration ratio because both Use the Henderson-Hasselbach equation to calculate the new pH. (Note: You can use the species are in the same volume.) Now take a fresh 60 mL of the 0.50 M pH 6.83 buffer and add 3.7 mL of 1.00 M NaOH. Using steps similar to those above, calculate the new pH of the solution. c. 88 Buffers and Salt Hydrolysis Laboratory 6

Please answer all parts.

Answer 1

Sol .

(a) As Conc. of HCl = 1.00 M

Volume of HCl = 1.7 mL

So , Molar quantity of HCl = Conc. of HCl × Volume of HCl

= 1.00 × 1.7 = 1.7  mmol

As HCl is a strong acid , so , Molar quantity of HCl = Molar quantity of H3O+ = 1.7 mmol

After addition of H3O+ , molar quantity of H2PO4- increases and molar quantity of HPO42- decreases .

So , Final molar quantity of H2PO4-

= Initial molar quantity of H2PO4- +  molar quantity of H3O+ added  

= 11.8 +  1.7 = 13.5

And , Final Molar quantity of HPO42-  

= Initial molar quantity of HPO42- - molar quantity of H3O+ added

= 18.2 - 1.7 = 16.5

(b) Now ,

New HPO42- / H2PO4- molar quantity ratio    = 16.5 / 13.5 = 1.2222

As pKa = 6.64

So , Using Henderson - Hasselbalch equation ,

New pH = pKa + log ( HPO42- / H2PO4- )

= 6.64 + log ( 1.2222 )

= 6.73  

(c) Now , Molar quantity of NaOH added = Conc. of NaOH × Volume of NaOH = 1.00 × 3.7 = 3.7 mmol

As NaOH is a strong base ,

So , Molar quantity of OH- = Molar quantity of NaOH = 3.7 mmol

Now , Molar quantity of HPO42- increases and molar quantity of H2PO4- decreases .

So ,

Final Molar quantity of HPO42- = Initial Molar quantity of HPO42- + Molar quantity of OH- added  

= 18.2 + 3.7 = 21.9

and , Final molar quantity of H2PO4- = Initial Molar quantity of H2PO4- - Molar quantity of OH- added  

= 11.8 - 3.7 = 8.1

Now , Using Henderson - Hasselbalch equation ,

New pH = pKa + log ( HPO42- / H2PO4- )

= 6.64 + log ( 21.9 / 8.1 )

= 7.07