What molarity of a NH4NO3 solution has a pH of 5.220? Kb for NH3 is 1.8 x 10^-5.
Choose one answer.
a. 6.31x 10-6.
b. 2.21 x 10-6
c. 5.56x 10-10.
d. 0.065
I started by taking the inverse log of the PH to get (6.02*10^-6) then I was thinking its not the way I should start but perhaps.....using the Kb
1.00*10^-14/1.8*10^-5 which gave me 5.5*10^-10 so I am completely lost on which direction I need to take.....
I got the answer incorrect because I picked C thinking it was that but its not.
To find the molarity of the NH4NO3 solution with a pH of 5.220, you need to consider the hydrolysis of NH4NO3 and its effect on the pH.
NH4NO3 can be broken down into NH4+ and NO3- ions. The NH4+ ion is acidic and can hydrolyze to produce H3O+, which lowers the pH.
The hydrolysis reaction is as follows:
NH4+ + H2O ⇌ NH3 + H3O+
To find the molarity (M) of NH4NO3 that gives a pH of 5.220, you can use the expression for the hydrolysis constant (Kh) and the relationship between Kb and Kh:
Kh = Kw/Kb
where Kh is the hydrolysis constant, Kw is the ion product of water (1.0 x 10^-14), and Kb is the base ionization constant of NH3.
Rearranging the equation:
Kb = Kw/Kh
Given that Kb for NH3 is 1.8 x 10^-5, you can substitute the values into the equation above:
1.8 x 10^-5 = 1.0 x 10^-14/Kh
Solving for Kh:
Kh = 1.0 x 10^-14/(1.8 x 10^-5)
Kh ≈ 5.56 x 10^-10
Since NH4+ is acidic, it reacts with water to produce H3O+. Therefore, increasing the concentration of NH4NO3 increases the concentration of NH4+ and, subsequently, H3O+ ions.
To determine the molarity of the NH4NO3 solution, you need to use the relationship between Kh and the concentration of NH4+ ions (M):
Kh = [H3O+][NH3]/[NH4+]
Given that the concentration of NH3 and NH4+ are the same in the solution, let's call them both M.
Kh = [H3O+][M]/[M]
Kh = [H3O+]
5.56 x 10^-10 = [H3O+]
Now, take the negative logarithm of [H3O+] to get the pH:
-pH = -log[H3O+]
5.220 = -log([H3O+])
[H3O+] = 10^(-5.220)
[H3O+] ≈ 6.31 x 10^(-6)
Since the concentration of NH4+ ions (M) is equal to the concentration of H3O+ ions, the molarity of the NH4NO3 solution is approximately 6.31 x 10^(-6) M.
Therefore, the correct answer is a. 6.31 x 10^-6.
To find the molarity of the NH4NO3 solution, you need to consider the hydrolysis reaction of NH4NO3 in water:
NH4NO3 (aq) + H2O (l) -> NH4+ (aq) + NO3- (aq)
NH4+ acts as a weak acid, and it will react with water to produce NH3 (aq) and H3O+ (aq):
NH4+ (aq) + H2O (l) -> NH3 (aq) + H3O+ (aq)
Since NH4NO3 is a salt of a weak acid (NH4+) and a strong base (NO3-), you can use the expression for the equilibrium constant of the hydrolysis reaction to calculate the concentration of NH4+ ions in the solution:
Kw = Ka * Kb
where Kw is the ion-product constant of water (1.0 x 10^-14), Ka is the acid dissociation constant of NH4+, and Kb is the base dissociation constant of NH3.
Rearranging the equation, we get:
Ka = Kw / Kb
Substituting the given values, we have:
Ka = (1.0 x 10^-14) / (1.8 x 10^-5)
Ka = 5.56 x 10^-10
Now, you can calculate the concentration of NH4+ ions using the equation for the dissociation constant of a weak acid:
Ka = [NH3] * [H3O+] / [NH4+]
Since [H3O+] can be approximated as the same value as the concentration of NH4+, you can rewrite the equation as:
Ka = [NH3] * [NH4+] / [NH4+]
Ka = [NH3]
Now, you can get the concentration of NH3:
[NH3] = 5.56 x 10^-10 M
Since NH4NO3 dissociates into one NH4+ and one NO3- ion, the concentration of NH4NO3 is equal to the concentration of NH4+:
[NH4NO3] = [NH4+] = 5.56 x 10^-10 M
So, the correct answer is c. 5.56 x 10^-10.
You were going great guns--you just stopped too soon.
NH4^+ + HOH ==> NH3 + H3O^+
So your Ka is correct at 5.55 x 10^-10
Ka = (NH3)(H3O^+)/(NH4^+)
You have the (H3O^+) and (NH3) is the same as (H3O^+). solve for (NH4^+), the only unknown in the equation.