A solution is prepared by mixing 100 mL of 0.500 M NH3 and 100 mL of 0.500 M HCl. Assuming that the volumes are additive, what is the pH of the resulting mixture?
kb for ammonia is 1.8x10-5
The base and acid are exactly neutralized; the pH is determined solely by the hydrolysis of the salt.
(NH4Cl) = 50 mmols/200 mL = 0.25M.
......NH4^+ + H2O..> NH3 + H3O^+
I.....0.25.............0......0
C......-x..............x......x
E.....0.25-x...........x......x
Ka for NH4^+ = (Kw/Kb for NH3) = (x)(x)/(0.25-x)
Solve for x = (H3O^+) and convert that to pH.
To find the pH of the resulting mixture, we need to determine the concentration of the resulting solution.
First, let's calculate the number of moles of NH3 and HCl in the given solution:
Molarity (M) = Moles (mol) / Volume (L)
Given:
Molarity of NH3 (M_NH3) = 0.500 M
Volume of NH3 (V_NH3) = 100 mL = 0.1 L
So, the number of moles of NH3 (n_NH3) can be calculated as:
n_NH3 = M_NH3 * V_NH3
n_NH3 = 0.500 M * 0.1 L
n_NH3 = 0.05 mol
Similarly, the number of moles of HCl (n_HCl) can be calculated as:
n_HCl = M_HCl * V_HCl
n_HCl = 0.500 M * 0.1 L
n_HCl = 0.05 mol
Since NH3 and HCl react in a 1:1 ratio, the moles of NH3 and HCl would completely neutralize each other.
Now, let's determine the concentration (M) of NH3 and HCl in the resulting solution:
Volume of the resulting solution = Volume of NH3 + Volume of HCl
Volume of the resulting solution = 100 mL + 100 mL = 200 mL = 0.2 L
The concentration of NH3 (M_NH3_res) in the resulting solution is given by:
M_NH3_res = n_NH3 / Volume of the resulting solution
M_NH3_res = 0.05 mol / 0.2 L
M_NH3_res = 0.25 M
The concentration of HCl (M_HCl_res) in the resulting solution is also 0.25 M.
Now, given that the Kb for ammonia (NH3) is 1.8x10^-5, we can use this information to find the pOH of the resulting solution. The pOH will then be used to determine the pH.
Kb = [NH4+][OH-] / [NH3]
Since the solution is neutral, the concentration of [NH4+] is equal to [OH-]. Let's call this concentration x.
Kb = x^2 / (0.25 - x)
Since the Kb value is very small, it is safe to assume that x is much less than 0.25. This assumption allows us to simplify the equation to:
Kb ≈ x^2 / 0.25
Rearranging the equation gives us:
x^2 = Kb * 0.25
x = sqrt(Kb * 0.25)
Plugging in the values, we get:
x = sqrt(1.8x10^-5 * 0.25)
x = 0.006
Now, we can calculate the pOH of the solution:
pOH = -log10([OH-])
pOH = -log10(0.006)
pOH ≈ 2.22
Finally, we can calculate the pH of the resulting solution using the equation:
pH = 14 - pOH
pH = 14 - 2.22
pH ≈ 11.78
Therefore, the pH of the resulting mixture is approximately 11.78.