What is the percent purity of the calcium hydroxide in this titration problem?

[Benzoate]

Homework Statement

A 0.2600-g sample of impure Ca(OH)2 is dissolved in enough water to make 45.70 mL of solution. 20.00 mL of the resulting solution is then titrated with 0.2455-M HCl. What is the percent purity of the calcium hydroxide if the titration requires 9.88 mL of the acid to reach the endpoint?

Homework Equations

The Attempt at a Solution

1) determine the number of mmol of HCl

20.00 mL*(.2455 mmol/1 mL) = 4.91 mmol

2) Convert mmol HCl to OH^1- using the notion that H^+ + OH^1- => H2O

4.9 mmol H^+ *(1mmol OH^1-/1 mmol H^1+) = 4.9 mmol OH^1-

3) Convert the mmol of OH^1- in 45.70 mL of the solution of Ca(OH)2

4.9 mmol OH^1- * (1mmol Ca(OH)2/1mmol (OH)^1-) *(74.1 mg Ca(OH)2/1mmol Ca(OH)2) = 363.09 mg Ca(OH)2

4.) I used the amount of Ca(OH)2 in 9.88 mL of solution of mass of Ca(OH)2

363.09 mg Ca(OH)2 * ((1000 mL/9.88 mL)) = 36750 mg = 36.75 g Ca(OH)2

5.) Use the mass of Ca(OH)2 and mass of pellets to now determine percent purity

36.75 g Ca(OH)2 /.2600 * 100 = 14134.61538 %

Shouldn't the percent purity be smaller than 100 %

 

[kaisxuans]

i don't know much of titration but i do know an equation that may help you.
to solve for a Henderson-Hasselbalch equation:
pH=pKa+log(base/acid) (fraction)
maybe you can go study a little more on the subject. it will also help if you show us your own attempts to the questions.
(visit my blog!)

 

[Benzoate]

i don't know much of titration but i do know an equation that may help you.
to solve for a Henderson-Hasselbalch equation:
pH=pKa+log(base/acid) (fraction)
maybe you can go study a little more on the subject. it will also help if you show us your own attempts to the questions.
(visit my blog!)

those were my own attempts. There was a similar problem like the one in my homework in my textbook , and so I used their explanations of the steps of the problems because I thought it would make it more clear to the chemistry tutors

We have started on the Henderson -Hasselbalch equation yet.

 

[kaisxuans]

Oh i see
im really very sorry but i cannot help you in any other way. if you see my blog or profile you will know what i am...so i can't help you
SORRY
oh and you wrote wrong it is HAVENT

 

[Benzoate]

Oh i see
im really very sorry but i cannot help you in any other way. if you see my blog or profile you will know what i am...so i can't help you
SORRY

its okay . I appreciate you voluntary to help me .

 

[atavistic]

Use molarity equation m1v1=m2v2.First find molarity of Ca(OH)2 .You'll get the vol. of HCL used this way.

Hope i am correct.

 

[atavistic]

I am not sure about the above one but if u convert molarity into normality then using the concept that the no.of equivalents(of any compound) in any reaction remains the same you can get the answer.

 

[atavistic]

Molarity won't work since this is titration reaction.So here is my solution as it is tough to explain.If u have any problem in understanding please post.

Ca(OH)2 +2HCl -> CaCl2 + 2H20

Valence factor of Ca(OH)2 = 2

Valence factor of HCl = 1

Normality of HCl = 0.2455 = 0.2455N

Equivalents of HCl in 9.88ml = 0.2455 * 9.88 = 2.42554meq

This implies eq in 20ml impure Ca(OH)2 solution = 2.42554meq
Equivalents in 45.7ml = 5.54meq

This implies pure Ca(OH)2 in 0.26g sample has 5.54meq

Mass of 5.54meq = 5.54/1000 * Eq weight of Ca(OH)2 = 0.205g

Percent Purity= Pure/Impure *100

I guess you can work that out :D

 

[chemisttree]

Start at the beginning... write out the titration reaction.

 

[Borek]

4.9 mmol OH^1- * (1mmol Ca(OH)2/1mmol (OH)^1-) *(74.1 mg Ca(OH)2/1mmol Ca(OH)2) = 363.09 mg Ca(OH)2

1mmol/1mmol?

363.09 mg Ca(OH)2 * ((1000 mL/9.88 mL)) = 36750 mg = 36.75 g Ca(OH)2

Why 1000 mL? What was your Ca(OH)2 solution volume?

 

[Kushal]

as chemisttree said you should be writing the equation first. you will then be able to determine how many moles of acid reacts with how many moles of alkali/base.

then calculate the number of moles of acid which reacted. use a simple proportion to find out the number of moles of alkali which reacted.(use the equation for that)

this amount of alkali will be present in 20 mL of solution. but you made 45.7 mL solution. you should be calculating the number of moles of the alkali in this volume, using simple proportion.

now convert this amount in mole to mass in grams.

from the 0.2600 g of solid, only the above calculated mass of Ca(OH)2 is present.
do your % purity and you're home and dry!

 

[Kushal]

simple proportion is most often easiest way to solve titration problems... formulas tend to be too confusing and you can easily make mistakes...