Posted by Kyle on Monday, February 16, 2009 at 6:48pm.
First you need to determine the concentration of NH3 in the final solution.
NH3 + HOH ==> NH4^+ + OH^-
Kb = (NH4^+)(OH^-)/(NH3)
You will need to look up Kb, if it isn't given in the problem, then plug in for OH^- and NH4^+ the OH^- you obtain from a pH of 11.65. That means a pOH of 2.35 and a (OH^-) of about 0.00447 but you need to go through it exactly since I've estimated. (Also, you must determine if the NH3 to use is NH3 - OH or just NH3. The size of the numbers will help you determine that.
Second, you need to determine the molarity of the NH3 solution you have (that's the 6.8% solution).
It has a density of 0.97 g/mL; therefore, 1000 mL will have a mass of
0.97 g/mL x 1000 mL = 970 grams.
How much of that is NH3. That will be
970 grams x 0.068 = 66 g NH3.
How many moles is that. grams/molar mass = moles = 66/17 = approximately 4 M.
Now use M(soln 1) x L(soln 1) = M(soln 2) x L(soln 2).
Post your work if you get stuck. And note that I've estimated at each step along the say so you need to redo all of this but use the exactly values.
Ok here is my solution to it but i am constantly told that it is wrong.
d=0.97g/mL
Mass %= 6.8%
1000mL= 1L
d=m/v m=d X v
m= 0.97g/mL X 1000mL
m=970g
mNH3 = 970g X 0.068=65.96g of NH3
MM of NH3 = 14g/mol + 3 X 1g/mol=17g/mol
nNH3 = 65.96g/17g/mol
n=3.88mol
[NH3]= 3.88mol/L or M
pH=11.65 thus pOH=2.35 thus [OH-]=4.466X10^-3
V=0.625L
Now using the ICE table for NH3 + H2O =(equilibrium sign) NH4^+ + OH^-
Kb=1.80 X 10^-5
I 3.88 0 0
C -x +x +x
E 3.88-x x x
Kb= [NH4^+][OH^-]/[NH3]
1.80 X 10^-5= x^2/3.88-x
neglecting the x because Kb is very small( I hope i am right)
1.80 X 10^-5 X 3.88 = x^2
using calculator and solving I got
x=8.3570X10^-3
similary X is [OH-]= 8.3570X10^-3
Using c1v1=c2v2
v1=c2v2/c1
v1=4.466X10-3 X 0.625/8.3570X10^-3
v1= 0.334L
which is 334mL and is the final answer.
But why in the NAME OF ARHENIUS, is the my answer not right.
DR. BOBB or any chem genius could tell me if I am missing anything becuase I took half hour to type the solution online. That is the how serious I am about this question.
and forgot to mention, I need a volume of 625mL of pH of 11.65....differnt from the original question posted....