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InterviewSolution
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Which would be decrease the pH of `25 cm^(3)` of a 0.01 M solution of hydrochloric acidA. The addition of `25cm^(3)` 0.005 M hydrochloric acidB. The addition of `25cm^(3)` of 0.02 M hydrochloric acidC. The addition of magnesium metalD. None of these |
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Answer» Correct Answer - B Given concentration of HCl = 0.01 M `pH = -log 10^(-2) = 2` In option (b) Concentration of HCl = 0.02 M M.eq `= 0.02 xx 25 = .5 M`. Eq of HCl M.eq in given condition `= 0.01 xx 25 = 0.25` Total M.eq in solution `= 0.25 + 0.50 = 0.75 = (.75)/(25+25) = 1.5 xx 10^(-2)N` Then new pH will be `= -log 1.5 xx 10^(-2) = 2 - log 1.5` Which is less than 2 ltbr. So we canvery that that by adding `25cm^(3)` of 0.2 M HCl the pH of `25 cm^(3)` of 0.01 M decrease. |
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