Q 1 :    

Given below are two statements:

Statement (I) : A Buffer solution is the mixture of a salt and an acid or a base mixed in any particular quantities.

Statement (II) : Blood is naturally occurring buffer solution whose pH is maintained by H2CO3/HCO3- concentrations.

In the light of the above statements, choose the correct answer from the options given below.               [2024]

  • Both Statement I and Statement II are false

     

  • Statement I is true but Statement II is false

     

  • Statement I is false but Statement II is true

     

  • Both Statement I and Statement lI are true

     

(3)      

           Acidic buffer: solution of weak acid and its salt with strong base. 

           This buffer is effective in the range 0.1<[acid][salt]<10

           Basic buffer: solution of weak base and its salt with strong acid.

           This buffer is effective in the range 0.1<[base][salt]<10



Q 2 :    

The pH of an aqueous solution containing 1M benzoic acid (pKa=4.20) and 1M sodium benzoate is 4.5. The volume of benzoic acid solution in 300 mL of this buffer solution is ________ mL. (given: log 2 = 0.3)              [2024]



(100)

As total volume of mixture is 300mL, let V mL of 1M benzoic acid is mixed with (300-V) mL of 1M sodium benzoate solution.

Moles of benzoic acid (nacid) = Molarity×volume

                                                = 1M×Vml = Vmmol

Moles of sodium benzoate (nsalt) = Molarity×volume = 1M×(300 - V)mL = (300 - V) mmol

A mixture of benzoic acid (weak acid) and sodium benzoate is an acidic buffer. For acidic buffer:

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

4.5=4.20+log300-VV

log300-VV=0.3

300-VV=2

V=100 mL



Q 3 :    

Only litre buffer solution was prepared by adding 0.10 mol each of NH3 and NH4Cl in deionised water. The change in pH on addition of 0.05 mol of HCl to the above solution is ____ ×10-2, (Nearest integer)                          

(Given : pKb of NH3 = 4.745 and log10 3 = 0.477)                              [2025]



(48)

Before adding HCl:

pOH=pKb+log[NH4+][NH3]

pOH=4.745+log0.10.1=4.745

After adding 0.05 mol HCl:

NH3+HClNH4Clt=0 (mol)0.10.050.1t=∞ (mol)0.1-0.0500.1+0.05=0.05=0.15

pOHnew=pKb+log[NH4+]new[NH3]new

pOHnew=pKb+log0.150.05

pOHnew=4.745+log3=4.745+0.477

Change in pH = Change in pOH

=pOHnew-pOH

=(4.745+0.477)-4.745=0.477=47.7×10-2

(Nearest integer = 48)