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# Making ice chart help

wrote...
Posts: 12
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9 years ago
 Making ice chart help I did poorly on this chapter and the ice tables were one reason. Can anyone explain them in detail? I have these two questions! thanks1) A(g) + 2B(g) ---> 2C(g)If .02 mol A + 0.01 mol B are placed in a 2.00 L vessel at equilibrium B is found to be 0.0042M what is Kc?2) What is the pH of a solution that is 0.012M in HA + has a Ka of 3.2x10^-8 Read 319 times 2 Replies

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wrote...
Subject Expert
9 years ago
 Hi LowriderMach,[A] = .02/2 = .01M[B ]i = .01/2 = .005M[B ]eq = .0042M                   A(g)        +       2B(g)                2C(g)Initial            .01                 .005                        0Change        -x                    -2x                      +2xEquilibrium  .01-x              .0042                     2xFrom [B ], we can see that.005-2x = .0042.0008 = 2xx = .0004[A]eq = .01-.0004 = .0096M[C]eq = 2*.0004   = .0008Kc = [A]*[B ]^2/[C]^2Kc = .0096*.0042^2/.0008^2Kc = .26462)                        HA       +         H2O         A-   +     H3O+Initial              .012                                           0              ~0Change            -x                                           +x              +xEquilibrium   .012-x                                          x                xKa = x^2/(.012-x)If Ka is less than 10^-4, you can approximate that .012-x ~ .012Ka = x^2/.0123.2*10^-8 = x^2/.012x^2 = 3.8*10^-10x   = .00002 = 2*10^-5since x is the concentration of H3O+-log(x) = pH-log(2*10^-5) = 4.7Hope this helps,Laser
LowriderMach Author
wrote...
9 years ago
 Quote from: Laser_3 (9 years ago)Hi LowriderMach,[A] = .02/2 = .01M[B ]i = .01/2 = .005M[B ]eq = .0042M                   A(g)        +       2B(g)                2C(g)Initial            .01                 .005                        0Change        -x                    -2x                      +2xEquilibrium  .01-x              .0042                     2xFrom [B ], we can see that.005-2x = .0042.0008 = 2xx = .0004[A]eq = .01-.0004 = .0096M[C]eq = 2*.0004   = .0008Kc = [A]*[B ]^2/[C]^2Kc = .0096*.0042^2/.0008^2Kc = .26462)                        HA       +         H2O         A-   +     H3O+Initial              .012                                           0              ~0Change            -x                                           +x              +xEquilibrium   .012-x                                          x                xKa = x^2/(.012-x)If Ka is less than 10^-4, you can approximate that .012-x ~ .012Ka = x^2/.0123.2*10^-8 = x^2/.012x^2 = 3.8*10^-10x   = .00002 = 2*10^-5since x is the concentration of H3O+-log(x) = pH-log(2*10^-5) = 4.7Hope this helps,Laseryou rock! thanks for this