Question

50 gram of sample of Ca(OH)2 is dissolved in 50ml of 0.5N HCl solution. The excess of HCl was titrated with 0.3 N NaOH. The volume of NaOH used was 20cc. calculate the percentage purity of Ca(OH)2

Answer 1

HCl + NaOH --- > NaOH + H2O

20 cc of .3 M NaOH

I assume you mean M (MOLAR) not N (Newtons) !

One Mol NaOH = 23+16+1 = 40 grams

20 cc is .02 liter

.02 liters * .3 moles/liter = .006 moles of NaO

so

.006 moles of HCl were left over from the original reaction

How many moles of HCl did we start with?

.05 liters * .5 moles/liter = .025 moles

so we used .025 - .006 = .019 moles of HCl in the original reaction

Ca(OH)2 + 2 HCl = Ca(CL)2 + 2 H2O

I therefore used .019/2 moles of Ca(OH)2

= .0095 moles of Ca(OH)2

Ca(OH)2 ---> 40+2+32 = 74 gm/mol

74 * .0095 = .703 gm of pure Ca(OH)2 were present

(.703/50)100 = 1.4 %