Consider the titration of a 25.0?mL sample of 0.175M CH3NH2 with 0.145M HBr. Determine each of the following.
the pH at 6.0mL of added acid
chemical equation :
CH3NH2
+ HBr --------------
> CH3NH3Br
Before rxn 0.025 L *0.175
M 0.145 M *
0.006L
0
= 0.004375 mols
=0.00087 mols
Afterrxn
0.004375 mols -0.00087 mols
0
0.00087mols
= 0.003505 mols
= 0.003505 mols /( 0.025 + 0.005 ) L
C = 0.116 M
CH3NH2 is a weakbase with Kb = 4.38*10 ^ - 4 .
[OH-] = ? Kb * C
= ? 4.38*10 ^ - 4 . * 0.116 M
= 0.7127 * 10 ^ -2 M
pOH = - log ( 0.7127 * 10 ^ -2 )
= 2.14
pH = 14 - 2.14
= 11. 85 ==>11.9
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