Question #db69c

1 Answer
Michael
Nov 20, 2015

[OH^-]=0.32"mol/l"

Explanation:

pK_w=pOH+pH

:.14=pOH+13.5

:.pOH=14-13.5=0.5

:.-log[OH^-]=0.5

:.log[OH^-]=-0.5

10^(-0.5)=0.32

:.[OH^-]=0.32"mol/l"

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