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pH and [OH-] of Basic Solution

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[OH-] and the pH for each of the following strong base solutions:

-- 0.082 M KOH

-- 1.065 g of KOH in 500.0 mL of solution

-- 10.0 mL of 0.0105 M Ca(OH)2 diluted to 500.0 mL

-- a solution formed by mixing 10.0 mL of 0.015 M Ba(OH)2 with 40.0 mL of 7.5x10-3 M NaOH.

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Acid-Base Equilibria
I am unsure of how to calculate

[OH-] and the pH for each of the following strong base solutions:
The pOH is calculated as pOH = - log[OH-]
And pOH + pH = 14

-- 0.082 M KOH
In solution, KOH is a strong base and electrolyte, so
KOH (aq) → K+ (aq) + OH- (aq)
0.82 0.082
Thus, the concentration of ...

Solution Summary

The solution provides examples of calculating the [OH-] and pH of several basic solutions.

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