Compute values for [H] [OH] and pH of the following aqueous solutions. Tabulate the answers for comparison. Spell out the reasoning in each calculation
a) 0.010M HCL
b)0.0010M NaOH
c) 0.010 N Ba(OH)2
(strong electrolytes assumed to be 100% dissociated.
Mostly the H+ is what is confusing me
(H^+) from HCl = (HCl) since HCl is 100% ionized. Then (H^+)(OH^-) = Kw = 1E-14
That allows you to calculate OH^-, the pH = -log(H^+).
For NaOH, it is 100% ionized; therefore, OH^- = (NaOH). Then solve for H^+ and pH.
Ba(OH)2 is a strong electrolyte; therefore, [Ba(OH)2] = 0.01 N, that is 0.01/2 = 0.005 M. (OH^-) is 2x[Ba(OH)2]. Then convert to H^+ and pH.
I still am lost on how to find OH- Kw being 1 x10^-14 could leave any possibility for H
For #1. 0.01 M HCl.
So H^+ = 0.01 M and since
(H^+)(OH^-) = 1E-14, then
(OH^-) = 1E-14/0.01 = 1E-12 M
For #2.
(NaOH) = 0.001 M so (OH^-) = 0.001 M
Then (H^+)(OH^-) = 1E-14
and (H^+) = 1E-14/0.001 = 1E-11 M
I'm confused by your question.
If you know (H^+) OR (OH^-) you immediately know the other BECAUSE
(H^+)(OH^-) ALWAYS equals 1E-14 (at least in water solution that is).
To compute the values for [H], [OH], and pH of the given aqueous solutions, we need to understand the concept of pH and the dissociation of strong electrolytes.
1) 0.010M HCl:
HCl is a strong acid that dissociates completely in water. The balanced chemical equation for the dissociation of HCl is:
HCl(aq) -> H+(aq) + Cl-(aq)
Since it is a strong electrolyte, we can assume that 0.010M HCl will dissociate completely, resulting in 0.010M H+ ions in solution. Therefore, [H+] = 0.010M.
Water also self-ionizes to a small extent:
H2O(l) <-> H+(aq) + OH-(aq)
At equilibrium, the concentration of [H+] is equal to the concentration of [OH-]. Since HCl is an acid, it will increase the [H+] concentration, causing the [OH-] concentration to decrease.
Therefore, [OH-] = 10^(-14) / [H+].
Using this information, we can now calculate the pH:
pH = -log[H+]
2) 0.0010M NaOH:
NaOH is a strong base that dissociates completely in water. The balanced chemical equation for the dissociation of NaOH is:
NaOH(aq) -> Na+(aq) + OH-(aq)
Since it is a strong electrolyte, we can assume that 0.0010M NaOH will dissociate completely, resulting in 0.0010M OH- ions in solution. Therefore, [OH-] = 0.0010M.
Similarly to the previous example, water self-ionizes to a small extent:
H2O(l) <-> H+(aq) + OH-(aq)
At equilibrium, the concentration of [H+] is equal to the concentration of [OH-]. Since NaOH is a base, it will increase the [OH-] concentration, causing the [H+] concentration to decrease.
Therefore, [H+] = 10^(-14) / [OH-].
Using this information, we can now calculate the pH:
pOH = -log[OH-]
pH = 14 - pOH
3) 0.010N Ba(OH)2:
Ba(OH)2 is a strong base and, similar to NaOH, it dissociates completely in water. The balanced chemical equation for the dissociation of Ba(OH)2 is:
Ba(OH)2(aq) -> Ba2+(aq) + 2OH-(aq)
Since it is a strong electrolyte, we can assume that 0.010N Ba(OH)2 will dissociate completely, resulting in 0.020N OH- ions in solution. Therefore, [OH-] = 0.020M.
Again, water self-ionizes to a small extent:
H2O(l) <-> H+(aq) + OH-(aq)
At equilibrium, the concentration of [H+] is equal to the concentration of [OH-]. Since Ba(OH)2 is a base, it will increase the [OH-] concentration, causing the [H+] concentration to decrease.
Therefore, [H+] = 10^(-14) / [OH-].
Using this information, we can now calculate the pH:
pOH = -log[OH-]
pH = 14 - pOH
By applying these calculations to the given aqueous solutions, you can tabulate the answers and compare them. Remember that pH represents the acidity or basicity of a solution, with values below 7 being acidic, 7 being neutral, and above 7 being basic.