50 ml 0.1 koh is mixed with 50 ml 0.1 m hcooh ph of the final solution is
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Explanation:
n
KOH
=
1000
50×0.2
=0.01 mol
n
HCOOH
=
1000
40×0.5
=0.02 mol
Moles of salt form HCOOK=
1000
50×0.2
=0.01 mol
Moles of left acid =0.02−0.01=0.01mol
pH=pK
a
+log
Acid
Salt
pH=−logK
a
+log
0.01
0.01
pH=3.744
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