calculate the ph at the equivilance point for the following

calculate the ph at the equivilance point for the following titration: .25 M HCl versus .25 M methylamine. Ka of methylammonium is 2.3 x 10^-11

Solution

finally only salt will be preent

molarity of the salt=0.25/2

=0.125

so,

using the formula

pH=7+0.5pKa+0.5logC

where C is the concentration of the acid,

we get

pH=7+0.5*10.638+0.5*log(0.125)

=11.8975

 calculate the ph at the equivilance point for the following titration: .25 M HCl versus .25 M methylamine. Ka of methylammonium is 2.3 x 10^-11Solutionfinally

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