pOH = -log[√{(Kw/Ka)(val)(Mgaram)}]
pOH = -log[√{(10^-14/10^-5)(2)(0.1)}]
pOH = -log[√{(10^-9)(0.2)}]
pOH = -log[√{2x10^-10}]
pOH = -log[√2 x 10^-5]
pOH = 5 -log[√2]
pH = 14-(5-log√[2])
pH = 9 +log√2
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pOH = -log[√{(Kw/Ka)(val)(Mgaram)}]
pOH = -log[√{(10^-14/10^-5)(2)(0.1)}]
pOH = -log[√{(10^-9)(0.2)}]
pOH = -log[√{2x10^-10}]
pOH = -log[√2 x 10^-5]
pOH = 5 -log[√2]
pH = 14-(5-log√[2])
pH = 9 +log√2