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Q.Calculate the pH of 1.5×10−31.5 \times 10^{-3} M solution of Ba(OH)2Ba(OH)_2.

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Step 1. Ba(OH)2→Ba2++2OH−Ba(OH)_2 \rightarrow Ba^{2+}+2OH^-: as a strong base fully dissociated, [OH−]=2×1.5×10−3=3.0×10−3[OH^-]=2\times1.5\times10^{-3}=3.0\times10^{-3} M. …

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