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25 mL of 0.1 N H3PO4 is titrated against 0.1 N NaOH. The pH of the solution after addition of 62.5 mL of the base is :(Given : Ka1=1×10−3,Ka2=1×10−8 and Ka3=1×10−13) |
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Answer» 25 mL of 0.1 N H3PO4 is titrated against 0.1 N NaOH. The pH of the solution after addition of 62.5 mL of the base is : (Given : Ka1=1×10−3,Ka2=1×10−8 and Ka3=1×10−13) |
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