Making ice chart help

Hi LowriderMach,


[A] = .02/2 = .01M
[B ]i = .01/2 = .005M
[B ]eq = .0042M

                   A(g)        +       2B(g)                2C(g)
Initial            .01                 .005                        0
Change        -x                    -2x                      +2x
Equilibrium  .01-x              .0042                     2x

From [B ], we can see that
.005-2x = .0042
.0008 = 2x
x = .0004
[A]eq = .01-.0004 = .0096M
[C]eq = 2*.0004   = .0008
Kc = [A]*[B ]^2/[C]^2
Kc = .0096*.0042^2/.0008^2
Kc = .2646

2)
                        HA       +         H2O         A-   +     H3O+
Initial              .012                                           0              ~0
Change            -x                                           +x              +x
Equilibrium   .012-x                                          x                x

Ka = x^2/(.012-x)
If Ka is less than 10^-4, you can approximate that .012-x ~ .012
Ka = x^2/.012
3.2*10^-8 = x^2/.012
x^2 = 3.8*10^-10
x   = .00002 = 2*10^-5

since x is the concentration of H3O+
-log(x) = pH
-log(2*10^-5) = 4.7

Hope this helps,
Laser

you rock! thanks for this