Question

at 90 degree celcius the ph of solution 0.0001 M KOH solution will be

Answer 1

Answer:

answer

0.001 M KOH [OH - ] = 0.001 M pOH = - log10[OH - ] pOH = - log (10-3) = 3 pH = pKw - pOH But 90oC is above room temprature, pKw decreases pH<14 - 3 pH<11 ...

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