Question

How many moles of sodium propionate should be added to one litre of an aqueous solution containing 0.02 mole of propionic acid to obtain a buffer solution of ph 4.75 ? What will be the ph if 0.01 mole of hclis dissolved in the above buffer solution . Compare the last ph value with ph of 0.01 molar hcl solution . Given ka for propionic acid = 1.34 8 10-5?

Answer 1

The pH will be 4.23 after addition of HCl.

Explanation:

According to Henderson's equation,

pH= pKa + log[Salt][Acid]pH= pKa + log[Salt]/[Acid]

Now, pH of solution = 4.75

pKa= -log Ka= -log 1.34 x 10-5= 4.870

[Acid]= 0.02 mol/L

Therefore,

4.75=4.870 + log[Salt]/(0.02)

or,log[Salt]/(0.02)=4.75 − 4.870

or, log[Salt]/(0.02)=−0.12

or, [Salt]/(0.02)= Antilog(−0.12)=0.8869

or,[Salt]=0.8869×0.02

or,[Salt]=0.017 mol/L

No. of moles of sodium propionate= Conc. × volume = 0.017 moles

Step 2:

Now, 0.01 mole of HCl is added to the above buffer.  

Before adding HCl, concentration of H+ ion will be

C2H5COOH  ↔  H+  +  C2H5COO−

Ka= [H+][C2H5COO−]/[C2H5COOH ]

or, [H+]= 1.348×10−5×0.02/0.017= 1.58×10−5 mol/L

Step 3:

Now, concentration of [H+] will be,

Ka=[H+][C2H5COO−]/[C2H5COOH]

or,1.348×10−5=[H+](0.007)/(0.03)

or, [H+]=1.348×10−5×0.007/0.03=5.77×10−5

pH= −log[H+] = − log (5.77×10−5) = 4.23

The pH will be 4.23 after addition of HCl.

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Answer 2

Answer:

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