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Given pH of a solution A is 3 and it is mixed with another solution B having pH 2. If both mixed then resultant pH of the solution will be |
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Answer» 3.2 `[H^(+)]_(B) = 10^(-3)M`. pH of the solution B = 2 `[H^(+)]_(B) = 10^(-2)M` `[H^(+)]_(B) = 10^(-3) + 10^(-2) = 10^(-3) + 10 XX 10^(-3) = 11 xx 10^(-3)`. `pH = -log(11 xx 10^(-3)) = 3 - log 11` `= 3 - 1.04 = 1.95`. |
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