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Title: Making ice chart help Post by: LowriderMach on Dec 5, 2013 I did poorly on this chapter and the ice tables were one reason. Can anyone explain them in detail? I have these two questions! thanks
1) 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 Title: Re: Making ice chart help Post by: Laser_3 on Dec 6, 2013 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 Title: Re: Making ice chart help Post by: LowriderMach on Dec 9, 2013 Hi LowriderMach, you rock! thanks for this :D[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 |