Pls solve Q 168 Calculate the pH of a buffer solution prepared by dissolving 30 g of Na2CO3 in - Chemistry -

Question

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

Vartika Jain · Accepted Answer

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)

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

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
$H C {O_{3}}^{-} = 5.63 \times 10^{- 11} [\log (\frac{133}{150}) = - 0.05]$
(1) 8.197
(2) 9.197
(3) 10.197
(4) 11.197