I will show you the continuation of this problem below. Because the is also 2.13 * 10 -2 M, you can then go on to use this as the new initial concentration for your next equilibrium equation. The from the first ionization is 2.13 * 10 -2 M. From the question above your Ka 1 is 3.5 * 10 -4 Ka 1 =Ĭross out the 1.3 on the left side. Now construct the equilibrium equation from the equilibrium values you have. What would be the concentration of H + be at equilibrium?įrom the chemical equation for the first ionization of H 2 S. VIDEO Solving Ka, Kb ICE Tables Demonstrated Example 3 (polyprotonic acid) : If the initial concentration of H 2 S was 1.3M, and the Ka 1 is 3.5 * 10 -4 and the Ka 2 is 7.9 * 10 -9. If we use our pH formulas then we can convert that to a pOH. Kb 1 =Ĭross out the 1.5 on the left side. From the question above your Kb is 4.3 * 10 -5. Then set up the ICE table and fill in the information from the problem (I only included aqueous because liquids don’t count for equilibrium). What would be the pOH be at equilibrium? Use the pH formulas when you need them.įrom the chemical equation for the ionization of NH 3 VIDEO Solving Ka, Kb ICE Tables Demonstrated Example 2 (with pH) : If the initial concentration of the base NH 3 was 1.5M, and the Kb is 4.3 * 10 -5. I have also removed the units to make visualization easier. We can make an assumption to get rid of the -x in. Then add the initial and change together to get the equilibrium concentration. Next make an X value representing the change in concentration when it shifts to the equilibrium.
Then set up the ICE table and fill in the information from the problem. What would be the concentration of OH – be at equilibrium?
VIDEO Solving Ka, Kb ICE Tables Demonstrated Example 1 : If the initial concentration of the base NH 3 was 0.7M, and the Kb is 4.3 * 10 -5. I will show one example a of a base with a single ionization and then one example of an acid with multiple ionizations. They can, howeve r, be confusing when you have more than one ionization with a polyprotonic acid. ICE tables for Ka and Kb work the same as they do for any other equilibrium. How do you use Ka or Kb in ICE or RICE tables?
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