Question

Veronal, a barbiturate drug, is generally administered as its sodium salt. What is the pH of a solution of

   NaC₈H₁₁N₂O₃ that contains 10. mg of the drug in 250 mL of solution? For veronal, HC₈H₁₁N₂O₃, the value

   of K_a is 3.7 x 10^-8.

Answer 1

mass =10.0mg

volume=250ml =0.250L

Molar mass of NaC₈H₁₁N₂O₃ =206.17 gram/mole

number of moles of salt = 10.0/206.17 =0.0485 moles

Concentration of salt = number of moles/volume in L

Concentration of salt = 0.0485/0.250 =0.194M

Concentration of salt =0.194M

NaC₈H₁₁N₂O3 ------------------ Na+ + C₈H₁₁N₂O3^-1

C₈H₁₁N₂O3^-1 + H2O ------------------- HC₈H₁₁N₂O3 + OH-

0.194                                                       0                   0

-x                                                             +x                  +x

0.194-x                                                      +x                +x

Kb = [HC₈H₁₁N₂O3][OH-]/[C₈H₁₁N₂O3-1 ]

KaxKb= Kw                  where Kw= 1.0x10^-14

Kb= Kw/Ka= 1.0x10^-14/3.7x10^-8=2.70x10^-7

Kb= 2.7x10^-7

2.7x10^-7 = x*x/(0.194-x)

for solving the equation

x=0.000229

[OH-]=0.000229M

-log[OH-]=-log(0.000229)

POH= 3.64

PH+POH=14

PH=14-POH

PH=14-3.64

PH=10.36