# chemistry

A 25.00 mL sample of 0.4 M dimethylamine (CH3)2NH is titrated with 0.150 M HCl.

What is the pH at the half-equivalence point?

Not sure what to do here exactly, I'm bad with the titration problems.

I know there are normally a couple tables (finding the number of moles at the equivalence point, using this to find concentrations, etc.) but with titrations I never can seem to get the reactions right.

Also I don't rememer the relevance of the half equilvance point. Is it where pH=pKa? I know there's some relevance to the half equivalence point, I just don't recall what it is.

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1. The titration equation is
(CH3)2NH + HCl ==> (CH3)2NH2^+ + Cl^-
Where is the equivalence point? It is
(25.00 x 0.4/0.15) = 66.67 mL so the total volume at the equivalence point will be 66.67 + 25.00 = 91.67 mL. The pH will be determined by the hydrolysis of the salt at the equivalence. The hydrolysis equation is
(CH3)2NH2^+ + H2O ==> (CH3)2NH + H3O^+
Set up an ICE chart.
(CH3)2NH2^+ will be 25.00*0.4000/91.67
(CH3)2NH = (H3O^+) = x
Solve for H3O^+ and convert to pH.

Yes, at the half way point the pH = pKa for an acid. Remember this is a base you are titrating so you want to use pKa for the base and not pKb.

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