What is the original molarity of an aqueous solution of ammonia (NH3) whose pH is 11.10 at 25 degrees C? (Kb for NH3=1.8 x 10^-5)
To find the original molarity of an aqueous solution of ammonia (NH3) at a given pH, we can use the relationship between pH, pOH, and the dissociation constant (Kw) of water.
1. Start by finding the pOH of the solution:
pOH = 14 - pH
pOH = 14 - 11.10
pOH = 2.90
2. Calculate the concentration of hydroxide ions (OH-) using the pOH:
pOH = -log[OH-]
[OH-] = 10^(-pOH)
[OH-] = 10^(-2.90)
3. Calculate the concentration of ammonia (NH3) using the concentration of hydroxide ions:
The reaction for the dissociation of ammonia (NH3) in water is:
NH3 + H2O -> NH4+ + OH-
Since the concentration of hydroxide ions is equal to the concentration of ammonium ions (NH4+), the concentration of ammonia can be assumed to be equal to the concentration of OH-.
Molarity of ammonia = [NH3] = [OH-] = 10^(-2.90)
Therefore, the original molarity of the aqueous solution of ammonia at pH 11.10 is 10^(-2.90). Please note that this value is in mol/L.
To find the original molarity of the aqueous ammonia solution, we need to use the relationship between pH, pOH, and the concentration of hydroxide ions (OH-) in the solution.
Since we are given the pH value of the solution, we can calculate the pOH value using the formula:
pOH = 14 - pH
In this case, pOH = 14 - 11.10 = 2.90.
The pOH is related to the concentration of hydroxide ions (OH-) in the solution by the equation:
pOH = -log[OH-]
To find [OH-], we will need to convert pOH back to a regular concentration value:
[OH-] = 10^(-pOH)
[OH-] = 10^(-2.90)
Now, we can determine the concentration of ammonia (NH3) using the equilibrium expression for the base hydrolysis reaction:
NH3 + H2O ⇌ NH4+ + OH-
The Kb expression for this reaction is:
Kb = [NH4+][OH-] / [NH3]
We can assume that the concentration of NH3 will be equal to the concentration of NH4+ since strong acids/bases or salts are not present.
Given that Kb for NH3 is 1.8 x 10^(-5), and we have found the concentration of OH- to be 10^(-2.90), we can now determine the concentration of NH3:
1.8 x 10^(-5) = (10^(-2.90))(x) / x
Simplifying the equation, we get:
1.8 x 10^(-5) = 10^(-2.90)
To solve this equation, we can take the logarithm of both sides using the common logarithm (log base 10) function:
log(1.8 x 10^(-5)) = log(10^(-2.90))
log(1.8) + log(10^(-5)) = -2.90
0.2553 - 5 = -2.90
-4.7447 = -2.90
Finally, we can take the antilog of both sides of the equation to solve for x, the concentration of NH3:
10^(-4.7447) = 10^(-2.90)
x = 10^(-4.7447)
So, the original molarity of the aqueous ammonia solution is approximately equal to 10^(-4.7447) M.
NH3 + H2O ==> NH4^+ + OH^-
Kb = (NH4^+)(OH^-)/(NH3) = 1.8E-5
Convert pH 11.10 to pOH, then to OH, substitute into Kb expression for NH4^+ and OH^-, and solve for (NH3), the only unknown in the equation.