calculate the OH- concentration in (a) 1.0M NH3 and (b) a solution that is 1.0M in NH3 and also 1.0M in NH4Cl. (Kb = 1.8 x 10^-5 for NH3)
thank you!
To calculate the OH- concentration in a solution, we need to first determine the concentration of the corresponding base and use the base dissociation constant. Here's how you can calculate the OH- concentration in the given solutions:
(a) 1.0M NH3:
NH3 (ammonia) is a weak base that reacts with water to produce OH-. The balanced equation for this reaction is:
NH3 + H2O ⇌ NH4+ + OH-
The base dissociation constant for NH3 (Kb) is given as 1.8 x 10^-5. To determine the OH- concentration, we need to set up an ICE table:
NH3 + H2O ⇌ NH4+ + OH-
I 1.0M -- 0M 0M
C -x -- x x
E 1.0-x -- x x
Using the Kb expression for NH3, we have:
Kb = ([NH4+][OH-]) / [NH3]
1.8 x 10^-5 = (x)(x) / (1.0 - x)
Since x is very small compared to 1.0, we can assume that 1.0 - x ≈ 1.0.
1.8 x 10^-5 ≈ x^2
x ≈ √(1.8 x 10^-5)
Therefore, the OH- concentration in 1.0M NH3 is approximately equal to √(1.8 x 10^-5) M.
(b) A solution that is 1.0M in NH3 and also 1.0M in NH4Cl:
In this case, NH3 is combined with NH4Cl, a salt of NH4+. When NH4Cl dissolves in water, it dissociates into NH4+ and Cl-. However, NH4+ acts as a weak acid in water and reacts with OH- to regenerate NH3. So, NH4+ consumes OH- and affects the OH- concentration.
Since NH4Cl is a strong electrolyte and completely ionizes, we have:
NH4Cl(s) → NH4+(aq) + Cl-(aq)
NH4+ + OH- ⇌ NH3 + H2O
In this case, we need to consider the common ion effect, where the NH4+ concentration will affect the OH- concentration. However, because the NH4Cl concentration is the same as the NH3 concentration (1.0M), the concentration of NH4+ will also be 1.0M.
Hence, to calculate the OH- concentration in a solution that is 1.0M in NH3 and 1.0M in NH4Cl, you would need to first calculate the concentration of NH4+, which is 1.0M, and then use the Kb value of NH3 (1.8 x 10^-5) as shown in part (a).
To calculate the OH- concentration, we need to consider the ionization of NH3 and the equilibrium reaction it forms with water.
(a) For a 1.0 M NH3 solution:
1. Write the equation for the ionization of NH3:
NH3 + H2O ⇌ NH4+ + OH-
2. Use the Kb expression for NH3 to calculate the OH- concentration.
Kb = [NH4+][OH-] / [NH3]
Rearrange the equation:
[OH-] = (Kb * [NH3]) / [NH4+]
The concentration of NH4+ is negligible in comparison to NH3 because NH4Cl is not present in this solution. Hence, [NH4+] is considered to be effectively zero.
[OH-] = (Kb * [NH3])
Substitute the given Kb value and NH3 concentration:
[OH-] = (1.8 x 10^-5) * (1.0)
[OH-] = 1.8 x 10^-5 M
So, the OH- concentration in a 1.0 M NH3 solution is 1.8 x 10^-5 M.
(b) For a solution that is 1.0 M in NH3 and 1.0 M in NH4Cl:
1. Write the equation for the reaction of NH3 with NH4Cl:
NH3 + H2O ⇌ NH4+ + OH-
NH4Cl ⇌ NH4+ + Cl-
2. The NH4+ concentration comes from the dissociation of NH4Cl:
[NH4+] = 1.0 M
3. The NH3 concentration remains the same:
[NH3] = 1.0 M
4. Use the Kb expression for NH3 to calculate the OH- concentration:
Kb = [NH4+][OH-] / [NH3]
Rearrange the equation:
[OH-] = (Kb * [NH3]) / [NH4+]
Substitute the given Kb value and NH3 and NH4+ concentrations:
[OH-] = (1.8 x 10^-5 * 1.0) / 1.0
[OH-] = 1.8 x 10^-5 M
So, the OH- concentration in a solution that is 1.0 M in NH3 and 1.0 M in NH4Cl is also 1.8 x 10^-5 M.
NH3 + H2O ==> NH4+ + OH^-
Kb = (NH4^+)(OH^-)/(NH3)
(NH4^+) = x
(OH^-) = x
(NH4) = 1.0M
Solve for x = (OH^-)
b) Same set up but (NH4^+) = x + 1.0
(OH^-) = x
(NH3) = 1.0-x
This will be a quadratic but you can simplify it by assuming x is so small as to be negligible so x+1.0 = 1.0 and 1.0-x = 1.0. I don't think this will make much difference in the answer but it makes it a bit easier to solve.
Of if you've had the Henderson-Hasselbalch equation you may use that since the NH4Cl/NH3 is a buffered solution.