Pls solve
Q.168. Calculate the pH of a buffer solution prepared by dissolving 30 g of N a 2 C O 3 in 500 mL of an aqueous solution containing 150 mL of 1 M HCl. Ka for
H C O 3 - = 5 . 63 × 10 - 11 log 133 150 = - 0 . 05
(1) 8.197
(2) 9.197
(3) 10.197
(4) 11.197

Dear Student,

Here first neutralisation reaction between Na2CO3 and HCl takes place. Na2CO3 +HCl  NaHCO3 +NaClBuffer is due to Na2CO3-NaHCO3.We know that, Equivalent of Na2CO3 = Equivalent of HCl Equivalent of HCl = NV=1×0.150=0.150    (n-factor fornHCL is 1, so M=N)and, Eqiuivalent of Na2CO3 =MassEquivalent mass=30106=0.283    (n-factor for Na2CO3 is also 1)So, after neutralisation, amount of Na2CO3 left =0.283-0.150=0.133And, 0.150 equivalent of Na2CO3 is converted to NaHCO3So, Equivalent of NaHCO3 = 0.150 According to Henderson Haselbach equation, pH = pKa + log [CO32-][HCO3-]      =-log(5.63×10-11)+log (0.133)(0.150)      =(-0.7505+11)+log 133150      =10.24+(-0.05)      =10.19Hence, correct answer is (3)

 

  • 13
What are you looking for?