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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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micike micike
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11 years ago
Calculate pH at the the point on the titration curve where a 50.00ml sample 0.0500M methylamine is titrated by addition of 0.0500 M HCl

1. 45.00 ml of HCl has been added


2. 51 mL of HCl has been added

please explain any steps if you know how to do it
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#1
11 years ago
Kb for CH3NH2 = 4.4 x 10^-4 ( Check this value on your book)
pKb =3.36
moles CH3NH2 = 0.0500 L x 0.0500 M=0.00250

moles HCl = 0.0450 L x 0.0500 M=0.00225
moles CH3NH2 in excess = 0.00250 - 0.00225 = 0.000250
moles CH3NH3+ formed = 0.00225
pOH = 3.36 + log 0.00225/ 0.000250=4.31
pH = 9.69

moles HCl = 0.051 L x 0.0500 M=0.00255
moles H+ in excess = 0.00255 - 0.00250=0.000050
total volume = 0.101 L
[H+]= 0.000050/ 0.101 =0.000495 M
pH = 3.31
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