Find the pH of the solution obtained when 32 mL of 0.087 M ethylamine is titrated to the equivalence point...?
I have no idea where to start. Do I need to use a rice/ice table for this problem?
This is my first time asking a question here, so...hopefully I'm doing all of this correctly! =)
Answer:
The formula for ethyl amine is C2H5NH2 and it is weak and monobasic. When it reacts with the acid, the cation formed is C2H5NH3+. The whole salt formula (cation and anion is C2H5NH3+Cl-, but we ignore the Cl-.
What we need to do is determine the molarity of the above salt of a weak base. First, we determine the moles of ethylamine that got titrated:
0.087 mol/L times 0.032 L = 0.002784 mol
Second, we need to determine the volume of acid added. Since we know ethylamine to be monobasic, we know there is a one to one ratio bewteen base and acid needed, so we need 0.002784 moles of HCl. How many mL is that?
0.002784 mol divided by 0.15 mol/L = 0.01856 L = 18.56 mL.
We now know the volume of the solution when titration is done: 32.00 mL + 18.56 mL = 50.56 mL. Notice that I added two sig figs onto the 32 mL. The equipment available for acid base titration allows measurements to this degree of precision. We need not round off to 50 mL. I'm going to carry an extra guard digits and round off at the end.
The molarity of the resulting solution is:
0.002784 mol divided by 0.05056 L = 0.05506 M
However, this is a salt of a weak baseand it undergoes hydrolysis (I'll use Et for C2H5):
EtNH3+ + H2O <===> EtNH2 + H3O+
The pH of the ending solution is going to be acidic.
There is another thing we need and this is not in the problem. It is the Ka for EtNH3+. This value can be looked up:
http://ifs.massey.ac.nz/resources/chemis...
I know it says Ka of a base, but trust me on this. The 1.56 x 10^-11 value is the one we need.
The Ka expression for the chemical equation above is as follows:
Ka = ([EtHN2] [H3O+]) divided by [EtNH3+]
We insert values as follows (you get them using some ICE table reasoning):
1.56 x 10^-11 = [(x) (x)] divided by (0.05506 - x)
Ignore the minus x to get this:
x = square root of [(1.56 x 10^-11) (0.05506)]
Take the negative logarithm of the resulting answer and there is you pH. Round it off to two sig figs and you're done.
I did the calculation and it's an acidic pH. All is well. Best wishes with your acid base unit.
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