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Chemistry 8 Online
OpenStudy (anonymous):

A volume of 15ml of HCl is titrated to an end point using 33.39ml of 0.05M Ca(OH)2 solution. Assuming complete neutralization, calculate the molarity of the HCl a. 0.056M b. 0.11M c. 0.22M d. None of these

OpenStudy (anonymous):

M1V1=M2V2 where M: Molarity and V: Volume Hence, M1=M2V2/V1 M1 = 0.05*33.39/15 = 0.11M Hence b.0.11

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