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10 ml of 0.1 M NaOH solution is diluted to 1000 ml solution. Proton concentration of the solution at 298 K will be
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Correct option is C
Dilution law :V1N1=V2N2 10(0.1)2=1000×N2 N2=2×10-3N; [OH-]=Normality of strog base ∴OH-=2×10-3 M ; Thus H+=10-142×10-3=5×10-12 M ;Talk to our academic expert!
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