50gm of a mixture of ca(OH)2 is dissolved in 50ml of 0.5N hcl solution, the excess of hcl was titrated with 0.3N -naoh. The volume of naoh used was 20cc. Calculate % purity of ca(OH)2

Question

The answer 41.35%

Who helps me step by step

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 %

Answer 2

Hey, If you label this as CHEMISTRY, you will get a chemistry teacher instead of a Physicist (me) or Mathematician (Steve).
Yours truly
Damon

Answer 3

In chemistry N stands for normality and in the case of NaOH the N and M have the same value.

Answer 4

Oh, thanks !