Question

Find out the PH of 10^-3 Molar NaoH...​

Answer 1

Explanation:

We examine the equilibrium:

2H2O(l)⇌H3O++−OH

Under standard conditions, [H3O+][HO−]=10−14.

And if we take log10 of both sides:

log10[H3O+]+log10[HO−]=−14

Om rearrangement:

14=−log10[H3O+]−log10[HO−]

But by definition, −log10[H3O+]=pH, and log10[HO−]=pOH.

And thus our definining relationship: pH+pOH=14.

Since (finally!) pOH=−log10[HO−]=−log10(1×10−3)=−(−3)