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find the ph of 0.100M NaC2H3O2 solution.

do i just write an equation
and solve for x

NaC2H3O2 +H2O= Na2O + HC2H3O2-
0.100 0 0
so x^2/0.100-x =1.8 X10^-5
x=1.3X10^-3
then do ph=-log(1.3X10-3)
ph=2.9
is this right?

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