1) find the hydronium ion concentration, the hydroxide ion concentration and the pH of a 0.000001 M solution of NaOH.
2) which of salts hydrolyze in aqueous solution : (NH4)2S, Na2So3,(CH3COO)2Ca?
Write the ionic equationss for the hydrolysis reacrions. Predict the pH of solution (pH=7 , pH>7 , ph<7) ?
The tendency for students is to say that the (OH^-) is 1E-6M but this is only 10 times more than is in H2O when it ionizes; therefore, we must add that in to the mixture. Here is how you do it.
NaOH is a strong electrolyte; therefore it will ionize 100% as follows.
..................NaOH --> Na^+ + OH^-
I.................1E-6..........0...........0
C..............-1E-6.........1E-6.....1E-6
E..................0............1E-6.....1E-6
Then add this to the ionization of water.
...............H2O ==> H^+ + OH^-
I..............liquid.......0........1E-6
C............liquid.......x...........x
E............liquid.......x.........1E-6+x
Kw = (H^+)(OH^-) = 1E-14
(x)(1E-6 + x) = 1E-14
Solve for x = (H^+), evaluate 1E-6+x for OH^- and convert all to pH.
For 2, look on Google and see what you can do for yourself.
Thank you very much!
To find the hydronium ion (H3O+) concentration, hydroxide ion (OH-) concentration, and pH of a solution, you'll need to use the concept of dissociation of water and the concentration of the given compound.
1) For a 0.000001 M solution of NaOH, you can assume complete dissociation of NaOH into Na+ and OH- ions.
1 NaOH → 1 Na+ + 1 OH-
Since NaOH is a strong base, it completely dissociates into Na+ and OH- ions. Therefore, the concentration of OH- ions in the solution is also 0.000001 M.
Since water autoionizes into H3O+ and OH- ions, we can use this information to calculate the hydronium ion concentration (H3O+). In water, the product of the concentrations of H3O+ and OH- is always constant at 1x10^-14 at 25 degrees Celsius.
[H3O+][OH-] = 1x10^-14
Substituting the known concentration of OH- (0.000001 M):
[H3O+][0.000001] = 1x10^-14
[H3O+] = 1x10^-14 / 0.000001
[H3O+] = 1x10^-8 M
Since pH is defined as the negative logarithm (base 10) of the hydronium ion concentration:
pH = -log [H3O+]
Substituting the concentration of H3O+ (1x10^-8 M):
pH = -log (1x10^-8)
pH = 8
Therefore, the hydronium ion concentration is 1x10^-8 M, the hydroxide ion concentration is 0.000001 M, and the pH of the solution is 8.
2) To determine if a salt hydrolyzes in aqueous solution, you need to consider the nature of the ions formed when the salt dissolves.
a) (NH4)2S:
The ammonium ion (NH4+) is derived from a weak base (NH3), while the sulfide ion (S2-) is derived from a strong acid (H2S). Since the ammonium ion is a weak acid, it undergoes hydrolysis in water. The reaction and the resulting ionic equation are as follows:
(NH4)2S + H2O → NH4OH + HS-
Ionic equation:
NH4+ + H2O → NH4OH + H+
The hydrolysis of NH4+ produces ammonium hydroxide (NH4OH) and H+ ions. Therefore, the pH of the solution will be acidic (pH<7).
b) Na2SO3:
Both the sodium ion (Na+) and the sulfite ion (SO3^(2-)) are derived from strong acid and base respectively, and they do not undergo hydrolysis in water. Hence, there will be no reaction and no hydrolysis equation. The pH of the solution will remain unchanged.
c) (CH3COO)2Ca:
The acetate ion (CH3COO-) is derived from a weak acid (acetic acid, CH3COOH), while the calcium ion (Ca2+) is derived from a strong base (Ca(OH)2).
(CH3COO)2Ca + H2O → Ca(OH)2 + 2CH3COOH
Ionic equation:
Ca2+ + H2O → Ca(OH)2 + 2H+
The hydrolysis of Ca2+ produces calcium hydroxide (Ca(OH)2) and H+ ions. Therefore, the pH of the solution will be acidic (pH<7).
In summary, for (NH4)2S and (CH3COO)2Ca, the hydrolysis reactions will produce acidic solutions (pH<7). For Na2SO3, there will be no hydrolysis and no change in pH.