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An aqueous solution contains 0.469 m ammonia. Calculate the ph of the solution after the addition of 1.83×10^-2 moles of hydroponic acid to 155 ml of this solution. The volume does not change.
Accepted Answer
Dissociation of Ammonia
The dissociation of ammonia in water can be represented by the following equilibrium:
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
The equilibrium constant, Kb, for this reaction is 1.8 x 10^-5.
Initial Concentration of Ammonia
The initial concentration of ammonia is 0.469 M.
Addition of Hydroponic Acid
The addition of 1.83 x 10^-2 moles of hydroponic acid to 155 mL of the ammonia solution will react with some of the ammonia to form ammonium ions.
The balanced chemical equation for this reaction is:
HCl(aq) + NH3(aq) ---→ NH4+(aq) + Cl-(aq)
Assuming that all of the hydroponic acid reacts with ammonia, the concentration of ammonia will decrease and the concentration of ammonium ions will increase.
New Concentration of Ammonia
The new concentration of ammonia can be calculated using the following equation:
Original moles of ammonia - moles of ammonia reacted = moles of ammonia remaining
0.469 M x 0.155 L - 1.83 x 10^-2 moles = 0.0725 moles of ammonia remaining
0.0725 moles / 0.155 L = 0.468 M
New Concentration of Ammonium Ions
The new concentration of ammonium ions can be calculated using the following equation:
Original moles of ammonium ions + moles of ammonium ions formed = moles of ammonium ions remaining
0 moles + 1.83 x 10^-2 moles = 1.83 x 10^-2 moles of ammonium ions
1.83 x 10^-2 moles / 0.155 L = 0.118 M
Calculating the pH
The pH of the solution can be calculated using the following equation:
pH = -log[H+]
To calculate the concentration of H+, we need to use the following equation:
Kw = [H+][OH-]
where Kw is the ion product constant for water, which is 1.0 x 10^-14.
Rearranging the equation, we get:
[H+] = Kw/[OH-]
We can calculate the concentration of OH- using the following equation:
Kb = [NH4+][OH-]/[NH3]
Rearranging the equation, we get:
[OH-] = Kb * [NH3]/[NH4+]
Substituting the given values, we get:
[OH-] = 1.8 x 10^-5 * 0.468 M / 0.118 M = 7.4 x 10^-6 M
Substituting the calculated value of [OH-] into the equation for [H+], we get:
[H+] = 1.0 x 10^-14 / 7.4 x 10^-6 M = 1.35 x 10^-9 M
Finally, substituting the calculated value of [H+] into the equation for pH, we get:
pH = -log(1.35 x 10^-9) = 8.87
The dissociation of ammonia in water can be represented by the following equilibrium:
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
The equilibrium constant, Kb, for this reaction is 1.8 x 10^-5.
Initial Concentration of Ammonia
The initial concentration of ammonia is 0.469 M.
Addition of Hydroponic Acid
The addition of 1.83 x 10^-2 moles of hydroponic acid to 155 mL of the ammonia solution will react with some of the ammonia to form ammonium ions.
The balanced chemical equation for this reaction is:
HCl(aq) + NH3(aq) ---→ NH4+(aq) + Cl-(aq)
Assuming that all of the hydroponic acid reacts with ammonia, the concentration of ammonia will decrease and the concentration of ammonium ions will increase.
New Concentration of Ammonia
The new concentration of ammonia can be calculated using the following equation:
Original moles of ammonia - moles of ammonia reacted = moles of ammonia remaining
0.469 M x 0.155 L - 1.83 x 10^-2 moles = 0.0725 moles of ammonia remaining
0.0725 moles / 0.155 L = 0.468 M
New Concentration of Ammonium Ions
The new concentration of ammonium ions can be calculated using the following equation:
Original moles of ammonium ions + moles of ammonium ions formed = moles of ammonium ions remaining
0 moles + 1.83 x 10^-2 moles = 1.83 x 10^-2 moles of ammonium ions
1.83 x 10^-2 moles / 0.155 L = 0.118 M
Calculating the pH
The pH of the solution can be calculated using the following equation:
pH = -log[H+]
To calculate the concentration of H+, we need to use the following equation:
Kw = [H+][OH-]
where Kw is the ion product constant for water, which is 1.0 x 10^-14.
Rearranging the equation, we get:
[H+] = Kw/[OH-]
We can calculate the concentration of OH- using the following equation:
Kb = [NH4+][OH-]/[NH3]
Rearranging the equation, we get:
[OH-] = Kb * [NH3]/[NH4+]
Substituting the given values, we get:
[OH-] = 1.8 x 10^-5 * 0.468 M / 0.118 M = 7.4 x 10^-6 M
Substituting the calculated value of [OH-] into the equation for [H+], we get:
[H+] = 1.0 x 10^-14 / 7.4 x 10^-6 M = 1.35 x 10^-9 M
Finally, substituting the calculated value of [H+] into the equation for pH, we get:
pH = -log(1.35 x 10^-9) = 8.87
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