WHAT CONCENTRATION OF AMMONIA WOULD HAVE A PH OF 11.80?
The answer in the book is 2.2M
However, my calculations are as follows:
pH = 11.80
pOH = 14-11.80 = 2.20
pOH ≠ [NH3]
[OH-] = 10^-(pOH) = 10^-2.20 = 6.31*10^(-3)
[OH-] = [NH3] because NH3 is a strong acid ∴ decomposition ratio is 1 to 1.
[NH3] = 6.31*10^(-3)
Where's the flaw in my calculation?
The answer in the book is 2.2M
However, my calculations are as follows:
pH = 11.80
pOH = 14-11.80 = 2.20
pOH ≠ [NH3]
[OH-] = 10^-(pOH) = 10^-2.20 = 6.31*10^(-3)
[OH-] = [NH3] because NH3 is a strong acid ∴ decomposition ratio is 1 to 1.
[NH3] = 6.31*10^(-3)
Where's the flaw in my calculation?
-
Your flaw is in your assumption of Ammonia.
It is actually a *WEAK* base. With a Kb value of 1.8*10^-5
With a reaction as follows:
NH3+H2O<=>NH4+ + OH-
1.8*10^-5=[OH-][NH4+]/[NH3]
Now, in this case, the concentration of NH4+ is equal to the concentration of OH- because of the base hydrolysis.
1.8*10^-5=((6.31*10^-3)^2)/[NH3]
Then divide the number on the right from both sides: you will get something like:
0.452078=1/[NH3]
Take the inverse of both sides:
[NH3]=2.212 or 2.2 to two significant figures.
*note:* technically you should subtract 6.318*10^-3 from the original NH3 concentration because that is getting consumed in the hydroloysis process, however, it falls outside of the 10^3 rule for assumptions (our kb has a 10^-5 order and our predicted molarity is in the 10^0 order, a 10^5 order difference) and can therefore be neglected in calculations.
It is actually a *WEAK* base. With a Kb value of 1.8*10^-5
With a reaction as follows:
NH3+H2O<=>NH4+ + OH-
1.8*10^-5=[OH-][NH4+]/[NH3]
Now, in this case, the concentration of NH4+ is equal to the concentration of OH- because of the base hydrolysis.
1.8*10^-5=((6.31*10^-3)^2)/[NH3]
Then divide the number on the right from both sides: you will get something like:
0.452078=1/[NH3]
Take the inverse of both sides:
[NH3]=2.212 or 2.2 to two significant figures.
*note:* technically you should subtract 6.318*10^-3 from the original NH3 concentration because that is getting consumed in the hydroloysis process, however, it falls outside of the 10^3 rule for assumptions (our kb has a 10^-5 order and our predicted molarity is in the 10^0 order, a 10^5 order difference) and can therefore be neglected in calculations.