What is pH of 0.1M solution of NH4Br, if the pKb of NH3 is 4.74?
To find the pH of a solution of NH4Br, we need to consider the hydrolysis of the salt and the dissociation of NH4+.
NH4Br is a salt formed from the conjugate acid NH4+ (ammonium ion) and the conjugate base Br- (bromide ion). Since NH4Br is a salt of a weak base (NH4+) and a strong acid (HBr), we can ignore the hydrolysis of the anion (Br-) and focus on the dissociation of the ammonium ion (NH4+).
The equation for the dissociation of NH4+ is as follows:
NH4+ + H2O ⇌ NH3 + H3O+
The equilibrium constant for this reaction, known as the Kb value, can be calculated using the pKb value of NH3:
Kb = 10^(-pKb)
Kb = 10^(-4.74)
Now, let's assume that 'x' is the concentration of NH4+ that dissociates. Then, the change in concentration of NH4+ is 'x', while the change in concentration of NH3 and H3O+ is also 'x'. The initial concentration of NH4+ is 0.1 M. Since the concentration of NH4+ decreases by 'x', the final concentration of NH4+ is (0.1 - x) M.
Using the Kb expression, we can set up an equation to solve for 'x':
Kb = [NH3][H3O+] / [NH4+]
10^(-4.74) = x * x / (0.1 - x)
Simplifying the equation and solving for 'x' will give us the concentration of NH3 and H3O+. The pH can then be calculated by taking the negative logarithm of the concentration of H3O+.
Note: This equation requires solving a quadratic equation, so it may be necessary to use the quadratic formula or an appropriate calculator to find the value of 'x' and determine the pH.
To find the pH of a solution of NH4Br, we first need to determine the concentration of OH- ions in the solution.
NH4Br is a salt formed by the reaction between NH3 (ammonia) and HBr (hydrobromic acid). When NH4Br dissolves in water, it dissociates into NH4+ cations and Br- anions. The NH4+ ion can react with water to form NH3 and H3O+:
NH4+ + H2O ↔ NH3 + H3O+
The NH3 can further react with water to form NH4+ and OH-:
NH3 + H2O ↔ NH4+ + OH-
Since NH4Br is a salt of NH4+ and the pKb of NH3 is given, we can use this information to determine the concentration of OH- ions in the solution.
The pKb is defined as the negative logarithm (base 10) of the equilibrium constant (Kb) for the reaction of the base (NH3) with water. In this case, Kb = Kw/Ka, where Kw is the ion product constant of water (1.0 x 10^-14) and Ka is the acid dissociation constant of NH4+.
Since NH4+ is the conjugate acid of NH3, we can write:
Ka x Kb = Kw
From the given pKb = 4.74, we can calculate Kb:
Kb = 10^(-pKb)
Kb = 10^(-4.74)
Kb = 1.96 x 10^(-5)
Now, we can use this Kb value to determine the concentration of OH- ions:
Kb = [NH4+][OH-]/[NH3]
1.96 x 10^(-5) = [NH4+][OH-]/[NH3]
Since the concentration of NH4+ and NH3 in the solution is equal, we can assume [NH4+] = [NH3]. Let's denote this concentration as x:
Kb = x * x / x
1.96 x 10^(-5) = x^2
x = sqrt(1.96 x 10^(-5))
x = 0.00443
Therefore, the concentration of OH- ions in the solution is 0.00443 M.
To find the pH of the solution, we can use the equation:
pOH = -log10([OH-])
pOH = -log10(0.00443)
pOH = 2.35
The pH of the solution can be obtained by subtracting the pOH from 14:
pH = 14 - pOH
pH = 14 - 2.35
pH ≈ 11.65
So, the pH of the 0.1 M NH4Br solution is approximately 11.65.
.......NH4^+ + HOH ==> NH3 + H3O^+
I......0.1..............0.....0
C.......-x..............x.....x
E.....0.1-x.............x......x
Ka for NH4^+ = (Kw/Kb for NH3) = (x)(x)/(NH4^+)
Substitute and solve for x = (H3O^+) and convert to pH.
note: If pKb = 4.74 then Kb = 1.8E-5