Question

10 3 attempts left Check my work Be sure to answer all parts. Determine the percent ionization of the followving solutions of formic acid at 25°c: points (a) 0.036 M Print (b) 3.7 x 104 M References x10 % Enter your answer in scientific notation.) (c) 1.25 M

Answer 1

(a) 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 K_a = 1.8x10^-4

          c = concentration = 0.036 M

Plug the values we get a = 0.0707

∴ % dissociation = 0.0707 x 100 = 7.07%

(b) when c = 3.7x10^-4M

Then a = 0.697

∴ % dissociation = 0.697 x 100 = 69.7%

(c) when c = 1.25 M

Then a = 0.012

∴ % dissociation = 0.012 x 100 = 1.2%