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Question #db69c

1 Answer

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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