Ammonia,is a weak base with a Ka value of 1.8x10^-5 .What is the pH of a 0.22M ammonia solution?
Ammonia,is a weak base with a Kb value of 1.8x10^-5 .What is the pH of a 0.22M ammonia solution?
sorry the 2nd question is the right one
NH3 + HOH ==> NH4^+ + OH^-
Kb = (NH^+)(OH^-)/(NH3)
Do an ICE chart and plug into the expression for Kb. I get pOH = about 2.7 and pH = about 11.3 but you need to go through and not estimate as I did.
ok thanks alot...
the PartB asks for:
What is the percent ionization of ammonia at this concentration?
To determine the pH of a solution of a weak base, such as ammonia (NH3), we need to consider its equilibrium reaction with water:
NH3 + H2O ⇌ NH4+ + OH-
Ammonia reacts with water to form ammonium ions (NH4+) and hydroxide ions (OH-). Since the Ka value is given as 1.8x10^-5, we can assume that the equilibrium constant (Kw) for water, which represents the concentration of hydronium ions (H3O+) and hydroxide ions (OH-), is equal to 1.0x10^-14 at 25°C.
The equation for the equilibrium constant (Kw) is:
Kw = [H3O+][OH-]
Since the concentration of OH- is equal to the concentration of NH4+ (since one OH- is produced for every NH4+), and we assume x as the concentration of NH4+ and OH-, we can simplify the equation to:
Kw = x^2
Rearranging the equation gives:
x = sqrt(Kw)
Now, we can plug in the value of Kw (1.0x10^-14) to calculate x:
x = sqrt(1.0x10^-14) = 1.0x10^-7
The concentration of OH- and NH4+ is now known, but we need to find the concentration of NH3 in the solution. The equation for NH3 dissociating into NH4+ is:
NH3 ⇌ NH4+ + OH-
Since x represents the concentration of NH4+, which is equal to the concentration of OH-, we can substitute x as the concentration of NH3. The total concentration of NH3 is given as 0.22M, so:
0.22 = x
Now, we can calculate the pOH of the solution by taking the negative logarithm (base 10) of x (concentration of OH-):
pOH = -log(x) = -log(1.0x10^-7)
Using calculator:
pOH = 7
Since pH + pOH = 14 (at 25°C), we can calculate the pH:
pH = 14 - pOH = 14 - 7 = 7
Therefore, the pH of a 0.22M ammonia solution is 7.