Question

The acidie ingredient in vinegar is acetic acid. The pll of vinegar is anound 2.4, and the molar conceneraition of acetic acid in vinegar is around 0.85 M. Based on this information, determine the valueo ionization constant, Kfor acetic acid acid 11pts 2. A solution of sulfuric acid (H SO4, 25.00 ml.) was I11 pts sodium hydroxide. What was the concentration of the sulfuric acid? Aggr@NCCUCHEMI200Spring201 SExam

Answer 1

1. Let a be the dissociation of the weak acid
                            HA <---> H ⁺ + A^-

initial conc.            c               0         0

change                -ca            +ca      +ca

Equb. conc.         c(1-a)          ca       ca

Dissociation constant , K_a = ca x ca / ( c(1-a)

                                         = c a² / (1-a)

In the case of weak acids a is very small so 1-a is taken as 1

So K_a = ca²

==> a = √ ( K_a / c )

Given pH= 2.4

pH = -log [H+] = 2.4

             [H+] = 10^-2.4 = 3.98x10^-3 M

             [H+] = ca = 3.98x10^-3 M

          c = concentration = 0.85 M

a = ca / c = (3.98x10^-3 M )/ 0.85 = 4.68x10^-3

So Ka = ca² = 0.85 x (4.68x10^-3 )²

         Ka = 1.86x10^-5

2. Number of moles of NaOH , n = Molarity x volume in L

                                                = 0.1020 M x 34.55 mL x10^-3L/mL

                                                = 3.52x10^-3 mole

The balanced reaction is H2SO4 + 2NaOH ---> Na2SO4 + 2H2O

2 moles of NaOH reacts with 1 mole of H2SO4

3.52x10^-3 moles of NaOH reacts with 3.52x10^-3 mole/2 = 1.76x10^-3 mole of H2SO4

So concentration of sulfuric acid = number of moles / volume in L

                                               = ( 1.76x10^-3 mole) / (25.00mL x10^-3 L/mL)

                                               = 0.0705 M