what is the ph of a solution having [OH] =2.75×10power-2
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pH=-log[H^+]
thnn p[oH]=-log[oH]
pOH=-log2.75x10^-2
=+2(log2.75+log10) =2x(0.4393+1)=2
x1.4393=2.8786
pH+pOH=14
PH=14-pOH=14-2.8786=11.12
PH(Approx)=11
thnn p[oH]=-log[oH]
pOH=-log2.75x10^-2
=+2(log2.75+log10) =2x(0.4393+1)=2
x1.4393=2.8786
pH+pOH=14
PH=14-pOH=14-2.8786=11.12
PH(Approx)=11
Kanagasabapathy:
ans crct r nt
yeah..
answer is 11.44
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