Ammonia (NH3) ionizes according to the following reaction:
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH–(aq)
The base dissociation constant for ammonia (NH3) is Kb = 1.8 × 10–5. Ammonia (NH3) also has a chloride salt, ammonium chloride (NH4Cl), which is soluble in water.
If 0.070 M of ammonia (NH3) and 0.035 M of its salt ammonium chloride (NH4Cl) are mixed in a solution, what is the pH of this solution?
I came up with an answer of 9.6, is this correct?
That looks ok to me.
Well, I must say you're pretty close! But let me just clarify a few things for you. The pH scale runs from 0 to 14, where 0 is acidic, 7 is neutral, and 14 is basic. So, a pH of 9.6 would indeed indicate that the solution is basic. However, it might not be the exact pH value of the solution you described.
To calculate the pH, we need to consider the ionization of NH3 and the dissociation of NH4Cl. In this case, NH4+ acts as an acid and ionizes in water to release H+ ions, while NH3 acts as a base by accepting H+ ions to form OH- ions.
The concentration of NH4+ due to NH4Cl is 0.035 M, and since it dissociates completely, we have 0.035 M of NH4+ and 0.035 M of Cl- ions.
Now, let's consider the ionization of NH3. The Kb value given helps us calculate the concentration of OH- ions produced when NH3 accepts H+ ions. Therefore, we find that [OH-] = Kb * [NH3] = (1.8 × 10–5) * (0.070).
The concentration of NH4+ acts as the source of H+ ions, so [H+] = [NH4+] = 0.035 M.
Using the equation for the equilibrium constant of water, Kw = [H+][OH-] = 1.0 × 10^-14, we can solve for [H+].
[H+]* (1.8 × 10–5) * (0.070) = 1.0 × 10^-14
Solving this equation will give you the concentration of [H+], from which you can calculate the pH. So go ahead, do the calculations, and you should get the correct pH value!
To find the pH of the solution, we need to determine the concentration of hydroxide ions (OH-) in the solution. From the given information, we know that ammonium chloride (NH4Cl) is a salt that dissociates in water to form ammonium (NH4+) and chloride (Cl-) ions.
Since NH4+ is the conjugate acid of NH3, it can react with water to form hydroxide ions (OH-). Therefore, we need to calculate the concentration of ammonium ions (NH4+) and use it to find the concentration of hydroxide ions (OH-).
Let's start by writing the equation for the reaction:
NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)
Since the concentration of NH4Cl is given as 0.035 M, the concentration of NH4+ is also 0.035 M.
Next, we need to convert the concentration of NH4+ to the concentration of NH3. We can assume that NH3 is the limiting reagent and that all of NH4+ reacts to form NH3.
Since the reaction is in equilibrium, we can use the reaction quotient, Qc, to determine the direction of the reaction:
Qc = [NH3][H3O+]/[NH4+]
When Qc < Kb, the reaction will proceed to the right (towards NH3), and when Qc > Kb, the reaction will proceed to the left (towards NH4+). Therefore, we need to compare Qc to Kb.
Kb = 1.8 × 10^-5
[NH3] = 0.070 M
[H3O+] = unknown
[NH4+] = 0.035 M
Qc = [NH3][H3O+]/[NH4+] = (0.070)([H3O+])/0.035 = 2[H3O+]
Since we are given that NH3 and NH4Cl are mixed in equal molar amounts, Qc = [H3O+]. Therefore, we can substitute Qc with [H3O+] in the equation:
[H3O+] = 2[H3O+]
[H3O+] = 2(0.070)
[H3O+] = 0.14 M
Now, we can calculate the concentration of hydroxide ions (OH-) using the expression for the self-ionization of water:
Kw = [H3O+][OH-] = (1.0 × 10^-14)
Given that [H3O+] = 0.14 M and Kw = (1.0 × 10^-14), we can rearrange the equation to solve for [OH-]:
[OH-] = Kw/[H3O+] = (1.0 × 10^-14)/0.14
[OH-] ≈ 7.14 × 10^-14 M
Finally, we can determine the pOH of the solution:
pOH = -log10([OH-]) = -log10(7.14 × 10^-14)
pOH ≈ 13.15
Since pH + pOH = 14, we can find the pH:
pH = 14 - pOH = 14 - 13.15
pH ≈ 0.85
Therefore, the pH of the solution is approximately 0.85. Your answer of 9.6 is incorrect.
To determine the pH of the solution, we need to consider its components: ammonia (NH3) and ammonium chloride (NH4Cl).
First, let's focus on the ammonia (NH3). Ammonia is a weak base that can react with water to form ammonium ion (NH4+) and hydroxide ion (OH-). The equilibrium expression for this reaction can be written as:
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)
The base dissociation constant for ammonia, Kb, is given as 1.8 × 10^-5. This constant relates to the equilibrium position of the reaction and helps determine the concentration of hydroxide ions formed by the reaction.
Now, let's consider the ammonium chloride (NH4Cl). When ammonium chloride is dissolved in water, it completely dissociates into ammonium ions (NH4+) and chloride ions (Cl-). Since NH4Cl is a salt, it does not affect the pH of the solution directly.
To solve this problem, we need to determine the concentration of hydroxide ions (OH-) formed from the reaction between ammonia and water. We can then use the hydroxide ion concentration to calculate the pOH of the solution and convert it to pH.
Given that the initial concentration of ammonia (NH3) is 0.070 M, we can assume that the reaction between NH3 and H2O has reached equilibrium. This allows us to use an ICE (Initial, Change, Equilibrium) table to calculate the equilibrium concentrations of NH4+ and OH-.
Let's denote x as the concentration of OH-. At equilibrium, the concentration of NH3 will decrease by x, while NH4+ and OH- will both increase by x. Therefore, the equilibrium concentrations can be written as:
[NH3] = 0.070 - x
[NH4+] = x
[OH-] = x
Using the Kb expression, we can set up the equilibrium expression:
Kb = [NH4+][OH-] / [NH3]
Substituting the concentrations:
1.8 × 10^-5 = (x * x) / (0.070 - x)
Since the concentration of NH3 is much larger than x, we can assume that 0.070 - x ≈ 0.070.
1.8 × 10^-5 ≈ (x * x) / 0.070
Rearranging and solving for x:
x * x = 1.8 × 10^-5 * 0.070
x ≈ √(1.8 × 10^-5 * 0.070)
Calculating this value, we find that x ≈ 7.93 x 10^-3 M.
Now that we have determined the concentration of hydroxide ions (OH-), we can calculate the pOH of the solution using the formula:
pOH = -log[OH-]
Plugging in the value of [OH-], we get:
pOH = -log(7.93 x 10^-3) ≈ 2.10
Finally, to find the pH of the solution, we can use the relation:
pH + pOH = 14
pH = 14 - pOH
pH ≈ 14 - 2.10
pH ≈ 11.9
Therefore, the pH of the solution is approximately 11.9, not 9.6 as you mentioned.