Concentration of 50 ml , 1.0 M ethylamine (base) = 50 × 1
= 50 mM
Concentration of 50 ml , 0.5 M HCl (acid) = 50 × 0.5
= 25 mM
Here concentration of acid neutralizes concentration of base. Hence concentration of base remaining = 50 (mM) - 25 (mM)
= 25 mM
Now pOH = - log(base)
= - log ( 25×10-3)
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