Calculate the ph of 0.01m solution of sodium ethanoate
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ph=-log(H+)
(H+)=molarity *no. of H+ ions
= 0.01*4
=(O.H)=10-m
p OH= -log 10 -4m
ph=4
(H+)=molarity *no. of H+ ions
= 0.01*4
=(O.H)=10-m
p OH= -log 10 -4m
ph=4
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