Question

Hydrofluoric acid, HF, has a K a of 6.8 × 10í. What are [H3O+ ], [Fí ], and [OHí ] in 0.790 M HF?

[H3O+] = (unrounded)

[F ] = M

[OHí ] = (Enter your answer in scientific notation.)

Answer 1

                            HF + H₂O <---> H₃O ⁺ + F^-

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 = 6.8x10^-4

          c = concentration = 0.790 M

Plug the values we get a = 0.0293

So [H₃O⁺] = ca = 0.790x0.0293 = 2.32x10^-2 M

We know that [H₃O⁺][OH^-] = 1.0x10^-14

                              [OH^-] = (1.0x10^-14 )/ [H₃O⁺]

                                      = (1.0x10^-14) / ( 2.32x10^-2 )

                                     = 4.31x10^-13 M