How do you use Ka or Kb in ICE or RICE tables?
ICE tables for Ka and Kb work the same as they do for any other equilibrium. They can, however, be confusing when you have more than one ionization with a polyprotonic acid. I will show one example a of a base with a single ionization and then one example of an acid with multiple ionizations. Use assumption where you can.
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}. What would be the concentration of OH^{–} be at equilibrium?
First step form the chemical equation.
NH_{3 (aq)} <—> NH_{4}^{+}_{(aq)} + OH^{–}_{(aq)}
Then set up the ICE table and fill in the information from the problem.
NH_{3 (aq)} <— | —> NH_{4}^{+ }_{(aq)} + OH^{–}_{(aq)} | |
INITIAL | 0.7M | 0M 0M |
CHANGE | ||
EQUILIBRIUM |
Next make an X value representing the change in concentration when it shifts to the equilibrium.
NH_{3 (aq)} <— | —> NH_{4}^{+ }_{(aq)} + OH^{–}_{(aq)} | |
INITIAL | 0.7M | 0M 0M |
CHANGE | – x | + x + x |
EQUILIBRIUM |
Then add the initial and change together to get the equilibrium concentration.
NH_{3 (aq)} <— | —> NH_{4}^{+ }_{(aq)} + OH^{–}_{(aq)} | |
INITIAL | 0.7M | 0M 0M |
CHANGE | – x | + x + x |
EQUILIBRIUM | 0.7M – x | x x |
Now construct the equilibrium equation from the equilibrium values you have.
Kb = | [ NH_{4}^{+} ] [ OH^{–} ] = | [ x ] [ x ] = | 4.3 * 10^{-5 } |
[ NH_{3} ] | [ 0.7M – x ] |
We can make an assumption to get rid of the -x in [ 0.7M – x ]. So it becomes [ 0.7 ]. I have also removed the units to make visualization easier.
[ x ] [ x ] = | 4.3 * 10^{-5 } |
[ 0.7 ] |
Multiply the top left together
x^{2} = | 4.3 * 10^{-5 } |
[ 0.7 ] |
Multiply both sides by 0.7
0.7 * x^{2} = | 4.3 * 10^{-5} * 0.7 |
[ 0.7 ] |
Cross out the 0.7 on the left side.
0.7 * x^{2} = | 4.3 * 10^{-5} * 0.7 |
[ 0.7 ] |
Simplify
x^{2} = | 4.3 * 10^{-5} * 0.7 |
1 |
Multiply the right side.
x^{2} = | 3.01 * 10^{-5} |
1 |
Take the square (2) root of both sides.
x = | 5.49 * 10^{-3} |
1 |
COMPLETE ANSWER:[ OH^{–}] = 5.49 * 10^{-3}M
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}. What would be the pOH be at equilibrium? Use the pH formulas when you need them.
From the chemical equation for the ionization of NH_{3}
NH_{3(aq)} + H_{2}O_{(l)} <—> NH_{4}^{+}_{(aq)} + OH^{–}_{(aq)}
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).
NH_{3(aq)} <— | —> NH_{4}^{+}_{(aq)} + OH^{–}_{(aq)} | |
INITIAL | 1.5M | 0M 0M |
CHANGE | ||
EQUILIBRIUM |
Next make an X value representing the change in concentration when it shifts to the equilibrium.
NH_{3(aq)} <— | —> NH_{4}^{+}_{(aq)} + OH^{–}_{(aq)} | |
INITIAL | 1.5M | 0M 0M |
CHANGE | -x | + x + x |
EQUILIBRIUM |
Then add the initial and change together to get the equilibrium concentration.
NH_{3(aq)} <— | —> NH_{4}^{+}_{(aq)} + OH^{–}_{(aq)} | |
INITIAL | 1.5M | 0M 0M |
CHANGE | -x | + x + x |
EQUILIBRIUM |
Then add the initial and change together to get the equilibrium concentration.
NH_{3(aq)} <— | —> NH_{4}^{+}_{(aq)} + OH^{–}_{(aq)} | |
INITIAL | 1.5M | 0M 0M |
CHANGE | -x | + x + x |
EQUILIBRIUM | 1.5M – x | x x |
Now construct the equilibrium equation from the equilibrium values you have. From the question above your Kb is 4.3 * 10^{-5}.
Kb_{1} = | [ NH_{4}^{+} ] [ OH^{–} ] = | [ x ] [ x ] = | 4.3 * 10^{-5 } |
[ NH_{3} ] | [ 1.5M – x ] |
We can make an assumption to get rid of the -x in [ 1.5M – x ]. So it becomes [ 1.5 ]. I have also removed the units to make visualization easier.
[ x ] [ x ] = | 4.3 * 10^{-5 } |
[ 1.5 ] |
Multiply the top left together
x^{2} = | 4.3 * 10^{-5 } |
[ 1.5 ] |
Multiply both sides by 1.5
1.5 * x^{2} = | 4.3 * 10^{-5 }* 1.5 |
[ 1.5 ] |
Cross out the 1.5 on the left side.
1.5 * x^{2} = | 4.3 * 10^{-5 }* 1.5 |
[ 1.5 ] |
Simplify
x^{2} = | 4.3 * 10^{-5 }* 1.5 |
1 |
Multiply the right side.
x^{2} = | 6.45 * 10^{-5 } |
1 |
Take the square (2) root of both sides.
x = | 8.03 * 10^{-3 } |
1 |
Now we know our [OH-] = 8.03 * 10-3. If we use our pH formulas then we can convert that to a pOH.
pOH = | – log [8.03 * 10^{-3}] |
pOH = | – ( – 2.09 ) |
pOH = | 2.09 |
COMPLETE ANSWER: pOH = 2.09
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}. What would be the concentration of H^{+} be at equilibrium?
From the chemical equation for the first ionization of H_{2}S.
H_{2}S_{(aq)} <—> HS^{–}_{(aq)} + H^{+}_{(aq)}
Then set up the ICE table and fill in the information from the problem.
H_{2}S_{ (aq)} <— | —> HS^{–}_{(aq)} + H^{+}_{(aq)} | |
INITIAL | 1.3M | 0M 0M |
CHANGE | ||
EQUILIBRIUM |
Next make an X value representing the change in concentration when it shifts to the equilibrium.
H_{2}S_{ (aq)} <— | —> HS^{–}_{(aq)} + H^{+}_{(aq)} | |
INITIAL | 1.3M | 0M 0M |
CHANGE | – x | + x + x |
EQUILIBRIUM |
Then add the initial and change together to get the equilibrium concentration.
H_{2}S_{ (aq)} <— | —> HS^{–}_{(aq)} + H^{+}_{(aq)} | |
INITIAL | 1.3M | 0M 0M |
CHANGE | – x | + x + x |
EQUILIBRIUM | 1.3M – x | x x |
Now construct the equilibrium equation from the equilibrium values you have. From the question above your Ka_{1} is 3.5 * 10^{-4}
Ka_{1} = | [ HS^{–} ] [ H^{+} ] = | [ x ] [ x ] = | 3.5 * 10^{-4 } |
[ H_{2}S ] | [ 1.3M – x ] |
We can make an assumption to get rid of the -x in [ 1.3M – x ]. So it becomes [ 1.3 ]. I have also removed the units to make visualization easier.
[ x ] [ x ] = | 3.5 * 10^{-4 } |
[ 1.3 ] |
Multiply the top left together
x^{2} = | 3.5 * 10^{-4 } |
[ 1.3 ] |
Multiply both sides by 1.3
1.3 * x^{2} = | 3.5 * 10^{-4} * 1.3 |
[ 1.3 ] |
Cross out the 1.3 on the left side.
1.3 * x^{2} = | 3.5 * 10^{-4} * 1.3 |
[ 1.3 ] |
Simplify
x^{2} = | 3.5 * 10^{-4} * 1.3 |
1 |
Multiply the right side.
x^{2} = | 4.55 * 10^{-4} |
1 |
Take the square (2) root of both sides.
x = | 2.13 * 10^{-2} |
1 |
The [ H^{+} ] from the first ionization is 2.13 * 10^{-2}M. Because the [ HS^{–} ] 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. I will show you the continuation of this problem below. However, as you will notice when I finish, most of the time only the first ionization of an acid contributes significantly to the [ H^{+} ]. That is to say, the first ionization contributes about 99% or more of the value for your overall or final [ H^{+} ]. So whenever I have to do equilibrium problems with polyprotonic acids in them I ONLY DO THE FIRST IONIZATION AND THEN GIVE THAT AS MY ANSWER. So the answer I would give for this problem is 2.13 * 10^{-2}M.
Form the second chemical equation for the ionization of HS^{–}.
HS^{–}_{(aq)} <—> S^{2-}_{(aq)} + H^{+}_{(aq)}
Then set up the ICE table and fill in the information from the problem.
HS^{–}_{(aq)} <— | —> S^{2}^{–}_{(aq)} + H^{+}_{(aq)} | |
INITIAL | 2.13 * 10^{-2} | 0M 0M |
CHANGE | ||
EQUILIBRIUM |
Next make an X value representing the change in concentration when it shifts to the equilibrium.
HS^{–}_{(aq)} <— | —> S^{2}^{–}_{(aq)} + H^{+}_{(aq)} | |
INITIAL | 2.13 * 10^{-2} | 0M 0M |
CHANGE | -x | +x +x |
EQUILIBRIUM |
Then add the initial and change together to get the equilibrium concentration.
HS^{–}_{(aq)} <— | —> S^{2}^{–}_{(aq)} + H^{+}_{(aq)} | |
INITIAL | 2.13 * 10^{-2} | 0M 0M |
CHANGE | -x | +x +x |
EQUILIBRIUM | 2.13 * 10^{-2} -x | x x |
Now construct the equilibrium equation from the equilibrium values you have. From the question above your Ka_{2} is 7.9 * 10^{-9}
Ka_{2} = | [ S^{2}^{–} ] [ H^{+} ] = | [ x ] [ x ] = | 7.9 * 10^{-9 } |
[ HS^{–} ] | [ 2.13 * 10^{-2} – x ] |
We can make an assumption to get rid of the -x in [ 2.13 * 10^{-2} – x ]. So it becomes [ 2.13 * 10^{-2} ]. I have also removed the units to make visualization easier.
[ x ] [ x ] = | 7.9 * 10^{-9 } |
[ 2.13 * 10^{-2 }] |
Multiply the top left together
x^{2} = | 7.9 * 10^{-9 } |
[ 2.13 * 10^{-2 }] |
Multiply both sides by 2.13 * 10^{-2}
2.13 * 10^{-2 }* x^{2} = | 7.9 * 10^{-9 }(2.13 * 10^{-2}) |
[ 2.13 * 10^{-2 }] |
Cross out the 2.13 * 10^{-2} on the left side.
2.13 * 10^{-2}* x^{2} = | 7.9 * 10^{-9 }(2.13 * 10^{-2}) |
[ 2.13 * 10^{-2 }] |
Simplify
x^{2} = | 7.9 * 10^{-9 }(2.13 * 10^{-2}) |
1 |
Multiply the right side.
x^{2} = | 1.68 * 10^{-10 } |
1 |
Take the square (2) root of both sides.
x = | 1.30 * 10^{-5 } |
1 |
Bring together the concentrations from the first ionization and the second ionization.
COMPLETE ANSWER: 2.13 * 10^{-2}M +1.30 * 10^{-5 }M = 2.13 * 10^{-2}M
The second ionization adds less than 1% difference so if I have 3 sig figs it does not even count.
PRACTICE PROBLEMS: Solve the Ka or Kb equilibrium ICE table problems below. Use your pH formulas if you need them.
If the initial concentration of the base F^{–} was 12M, and the Kb is 9.4 * 10^{-6}. What would be the concentration of OH^{–} be at equilibrium?
Answer: [OH^{–}] = 1.06 * 10^{-2}M
If the initial concentration of the acid NH_{4}^{+} was 4.2M, and the Kb is 1.6 * 10^{-11}. What would be the concentration of H^{+} be at equilibrium?
Answer: [H^{+}] = 5.12 * 10^{-2}M….use the Kb to get the Ka then solve for [H^{+}]
If the initial concentration of the BrO_{2}^{–} was 0.8M in water, and the Kb is 2.3 * 10^{-5}. What would be the pH be at equilibrium?
Answer: pH = 11.6 ….solve for [OH-] and then use the pH formulas for the pH
Carbonic acid is created with an initial concentration of 3M. If the carbonic acid has a Ka_{1} of 9.8 * 10^{-5} and a Ka_{2} of 6.1 * 10^{-12}. What would be the pOH be at equilibrium?
Answer: pOH = 12.2 ….solve for [H^{+}] and then use the pH formulas for the pOH
H_{3}PO_{4} was put into water at a concentration of 0.098M. The Ka_{1} is 1.3 * 10^{-3}, the Ka_{2} is 7.2 * 10^{-6} and the Ka_{3} is 5.6 * 10^{-9}. What would be the concentration of OH- be at equilibrium?
Answer: [OH^{–}] = 8.85 * 10^{-13}….….solve for [H^{+}] and then use the pH formulas for the [OH^{–}]