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Calculate pH of mixture of (400mL,(1)/(2...

Calculate `pH` of mixture of `(400mL,(1)/(200)M Ba(OH)_(2))+(400mL.(1)/(50)M HCl)+(200mL "of water")`

Text Solution

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`[H^(+)]_(f)=((0.4xx(1)/(50))-(0.4xx(1)/(200)xx2))/(0.4+0.4+0.2)=4xx10^(-3)M`. So `pH=3-2log2=2.4`
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