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Determine the pH of the solution that results from the addition of 20.00 mL of 0.01 M Ca `(OH)_(2) ` to 30.00 mL of 0.01 M HClA. `11.30`B. `10.53`C. `2.70`D. `8.35` |
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Answer» Correct Answer - A `20.0 mL "of" 0.01 M Ca (OH)_(2) "has" Ca(OH)_(2)` `=20xx0.01` millimoles = 0.2 millimole `:. OH^(-)` ions present `=2xx0.2` millimole `=0.4` millimole 30.0 mL of 0.01 M HCl has HCl `=30xx0.01` millimoles = 0.3 millimole `:. H^(+)` ions present = 0.3 millimole 0.3 millimole `H^(+)` will neutralize 0.3 millimoe of `OH^(-)`. Hence, `OH^(-)` ions present after mixing = 0.1 millimole Volume of solution = 20 +30 = 50 mL Hence, molarity of `OH^(-)` ions `=(0.1)/(50)=2xx10^(-3)M` `pOH = - log (2xx10^(-3))=3-0.30=2.70` `:. pH = 14 - 2.70 = 11.30` |
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