Refer to Equilibrium Constants. Calculate the pH of a solution resulting from the addition of 50.0 mL of 0.30 M HNO3 to 50.0 mL of 0.30 M NH3.
To calculate the pH of the resulting solution from the addition of HNO3 and NH3, we need to consider how these two compounds react with each other, HNO3 being an acid and NH3 being a base. The reaction between HNO3 and NH3 is as follows:
HNO3 + NH3 ⟶ NH4+ + NO3-
Since HNO3 is a strong acid and completely dissociates in water, we can assume that it will completely react with NH3. Therefore, in the 50.0 mL of 0.30 M HNO3, we have 0.015 moles of HNO3. Similarly, in the 50.0 mL of 0.30 M NH3, we have 0.015 moles of NH3.
Based on the stoichiometry of the reaction, since HNO3 and NH3 react in a 1:1 ratio, 0.015 moles of both HNO3 and NH3 will be consumed completely in the reaction.
Now, calculating the concentration of NH4+ (ammonium ion) in the resulting solution:
Since the volume of the solution is 100.0 mL (50.0 mL + 50.0 mL), the total number of moles of NH4+ in the solution is 0.015 moles.
Using the equation for the equilibrium constant (Keq) for the reaction:
Keq = [NH4+]/[NH3][HNO3]
we can rearrange it to solve for [NH4+]:
[NH4+] = Keq × [NH3] × [HNO3]
The value of Keq for the reaction is approximately 10^10.
Plugging in the values:
[NH4+] = (10^10) × (0.015 moles) × (0.015 moles)
[NH4+] ≈ 0.00225
Since [NH4+] is the concentration of NH4+ in moles per liter, we need to convert it to molarity.
The volume of the resulting solution is 100.0 mL, which is equal to 0.1 L.
[NH4+] = 0.00225 moles / 0.1 L
[NH4+] = 0.0225 M
Now, to calculate the pOH of the resulting solution, we can use the concentration of NH4+, which is 0.0225 M.
pOH = -log10([OH-])
We know that in the reaction, NH4+ is formed, which can act as a weak acid and donate a proton (H+), resulting in the production of OH- ions. Since NH4+ is a weak acid, we can assume that it dissociates partially. However, in this case, it can be considered as fully ionized.
Therefore, [OH-] = [NH4+]
pOH = -log10(0.0225)
pOH ≈ 1.65
To calculate the pH of the solution, we use the equation:
pH = 14 - pOH
pH = 14 - 1.65
pH ≈ 12.35
Therefore, the pH of the resulting solution is approximately 12.35.
To calculate the pH of a solution resulting from the addition of two solutions, we need to consider the reaction between the substances involved and their equilibrium constants.
In this case, we have a reaction between nitric acid (HNO3) and ammonia (NH3), which can be represented as follows:
HNO3 + NH3 ⇌ NH4+ + NO3-
The equilibrium constant for this reaction is known as the acid dissociation constant for ammonia (Kb), which represents the extent to which ammonia accepts a proton (H+) to form ammonium ion (NH4+).
First, let's calculate the concentration of NH4+ and NO3- ions by considering the initial concentrations and volumes of the solutions:
NH4+ concentration = (0.30 M NH3)
(NO3- concentration = 0.30 M HNO3)
Since the volumes of the solutions are equal, the final volume of the solution after mixing will be 100.0 mL. Hence, the initial concentration of NH4+ and NO3- is:
NH4+ concentration = (0.30 M NH3) × (50.0 mL / 100.0 mL) = 0.15 M
NO3- concentration = (0.30 M HNO3) × (50.0 mL / 100.0 mL) = 0.15 M
Next, we need to find the concentration of OH- ions produced by the reaction between NH4+ and NO3-. To do this, we will use the Kb value for ammonia, which is 1.8 × 10^-5.
Kb = [NH4+][OH-] / [NH3]
Assuming the concentration of [OH-] produced, we can calculate the concentration of [NH4+] and [NH3] using the equation:
[NH3] = [NH4+] + [OH-]
Now, let's solve the equation. Since [NH4+] and [NH3] concentrations are equal initially (0.15 M), we can substitute these values into the equation:
[NH3] = 0.15 M + [OH-]
Applying the Kb expression,
1.8 × 10^-5 = (0.15 M + [OH-]) × [OH-] / 0.15 M
Rearranging the equation, we get:
[OH-]^2 + 0.15 [OH-] - 1.8 × 10^-5 = 0
Now, we can solve this quadratic equation to find the concentration of OH- ions.
Using the quadratic formula, where a = 1, b = 0.15, and c = -1.8 × 10^-5, we get:
[OH-] = (-0.15 ± √(0.15^2 - 4 × 1 × -1.8 × 10^-5)) / (2 × 1)
Calculating the values inside the square root:
(0.15^2 - 4 × 1 × -1.8 × 10^-5) = 0.0225 + 7.2 × 10^-5 = 0.022722
Taking the square root:
√(0.022722) ≈ 0.1508
Simplifying further:
[OH-] = (-0.15 ± 0.1508) / 2
Now, we consider both solutions for [OH-]:
[OH-] = (-0.15 + 0.1508) / 2 ≈ 0.0004 M (ignore the negative solution)
Since [OH-] represents the concentration of hydroxide ions, and pH is defined as the negative logarithm of the hydrogen ion concentration ([H+]), we can calculate pH using the equation:
pH = 14 - log10 [H+]
In this case, due to the reaction between NH4+ and OH-, we know [H+] = [OH-], so pH = 14 - log10 [OH-].
Calculating pH:
pH = 14 - log10 (0.0004) ≈ 12.4
Therefore, the pH of the solution resulting from the addition of 50.0 mL of 0.30 M HNO3 to 50.0 mL of 0.30 M NH3 is approximately 12.4.
After addition, all of the NH3 becomes NH4^+ and the pH is determined by dissociation of H^+.
NH4^+ -----------------> NH3 + H^+
Calculate new molarity after addition of acid.
Molarity of NH4^+=(0.3M*0.050L)/(0.100L)=0.15M
I......0.15 M...................0............0
C......-X.............,.......+X..........+X
E.....0.15 M-X..............X...........X
Kw/Kb=Ka=[products]/[reactants]
Ka=[x][x]/[0.15M-x]
5% dissociation allows for simplification....
Ka=x^2/[0.15M]
X=√(Ka*0.15M)
pH=-log[x]
Answer with the correct number of significant figures and.check 5% dissociation.