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Please help me!!!!! A volume of 15ml of HCI is titrated to an end point using 33.39 ml of 0.05M Ca(OH)2 solution. Assuming complete neutralization, calculate the molarity of HCI. A) 0.056M B) 0.11M C)0.22M D) none of these.
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first write down the equation for the reaction ! \[Ca(OH)2 + 2HCl \rightarrow CaCl _{2} + 2H _{2}O\] Looking a this we can ay the mole ratio of Ca(OH)2 to HCl is 1:2 So the no.of moles of HCl would be half the no of moles of Ca(OH)2 right? We can find the no. of moles of Ca(OH)2 using the given data. Use C=n/V C=Concentration n- no. of moles V= Volume Then we can find the no of moles of HCl Molarity= no. of moles of HCl/ Volume(in liters ) right ?
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