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
The answer 41.35%
Who helps me step by step
In chemistry N stands for normality and in the case of NaOH the N and M have the same value.
To calculate the percent purity of Ca(OH)2 in the given mixture, follow these steps:
Step 1: Determine the moles of HCl used:
Given that the HCl solution is 0.5N (Normality), which means it contains 0.5 moles of HCl per liter (dm^3) of solution.
We are given that 50ml (0.05L) of this HCl solution is used.
Moles of HCl = Volume (in L) x Normality = 0.05L x 0.5N = 0.025 moles
Step 2: Determine the moles of NaOH used to neutralize the excess HCl:
Given that the NaOH solution used is 0.3N, which means it contains 0.3 moles of NaOH per liter (dm^3) of solution.
We are given that 20cc (0.02L) of this NaOH solution is used.
Moles of NaOH = Volume (in L) x Normality = 0.02L x 0.3N = 0.006 moles
Step 3: Calculate the excess moles of HCl:
Since the reaction between HCl and NaOH is 1:1 (as per the balanced chemical equation), the moles of NaOH used will also be equal to the moles of HCl reacted.
Hence, the excess moles of HCl can be calculated as: Excess Moles of HCl = Moles of HCl used - Moles of NaOH used
Excess Moles of HCl = 0.025 - 0.006 = 0.019 moles
Step 4: Calculate the moles of Ca(OH)2 used:
Since each mole of Ca(OH)2 reacts with 2 moles of HCl (as per the balanced chemical equation), we can calculate the moles of Ca(OH)2 as: Moles of Ca(OH)2 = Excess Moles of HCl / 2
Moles of Ca(OH)2 = 0.019 / 2 = 0.0095 moles
Step 5: Calculate the mass of Ca(OH)2:
The molar mass of Ca(OH)2 can be determined as follows:
Atomic mass of Ca = 40.08 g/mol
Atomic mass of O = 16.00 g/mol (there are 2 oxygen atoms)
Atomic mass of H = 1.01 g/mol (there are 2 hydrogen atoms)
Molar mass of Ca(OH)2 = (40.08 + 2 * 16.00 + 2 * 1.01) g/mol = 74.09 g/mol
Mass of Ca(OH)2 = Moles of Ca(OH)2 x Molar mass of Ca(OH)2
Mass of Ca(OH)2 = 0.0095 moles x 74.09 g/mol = 0.704 g
Step 6: Calculate the percent purity of Ca(OH)2:
The percent purity is the mass of pure Ca(OH)2 divided by the mass of the mixture, multiplied by 100.
Given that 50g of the mixture is dissolved, the percent purity can be calculated as:
Percent Purity of Ca(OH)2 = (Mass of Ca(OH)2 / Mass of Mixture) x 100
Percent Purity of Ca(OH)2 = (0.704g / 50g) x 100 = 1.408%
Therefore, the percent purity of Ca(OH)2 in the given mixture is approximately 1.41%.
Oh, thanks !
Hey, If you label this as CHEMISTRY, you will get a chemistry teacher instead of a Physicist (me) or Mathematician (Steve).
Yours truly
Damon
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 %