If the [h+] in a solution is 1 × 10–1 mol/l, what is the [oh–]?

You need to know the following equations:
pH=-log[H⁺]
pOH=-log[OH⁻]
pH+pOH=14
then you can find the pH by using pH=-log(1×10⁻¹)
pH=1
then you can find pOH by doing pOH=14-1
pOH=13
then you can find [OH] by using [OH⁻]=10^(-pOH)
[OH⁻]=1×10⁻¹³mol/L

You could also use the equation Kw=[OH⁻][H⁺] where Kw=10⁻¹⁴
10⁻¹⁴/10⁻¹=10⁻¹³mol/L

I hope this helps.  Let me know if anything is unclear.

Rate law for this reaction israte of the reaction = k[a]²[b].second order in a  means concentration of a in power of two, and  first order in b means concentration of b in power of one.

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