1- what is the percentage of sodium carbonate in a sample of 0.3215 g. If 27.92 ml of 0.1131 M HCl were needed to titrate the sample




2- A 1.5 g of soda ash sample was dissolved in 250 mL water. Titration of 10 mL of the solution required 4 mL of 0.2 M HCl. What is the percent purity of soda ash ?



Could anybody help me ?

1.
a. Write the balanced equation of the reaction.
b. Find the number of moles of HCl in the volume given.
c. Using the mole ratio from the equation, find the number of moles of Na2CO3.
d. Find the weight of the number of moles of Na2CO3 you obtained in c. using the relative molecular mass of Na2CO3.
e. Find the percentage Na2CO3 in the sample, by using the fraction of mass you obtained in d. and the mass of the given sample.

2. The same principle applies. Write your equation, find the number of moles of acid, then in the 10 mL of solution. Find the number of moles in 250 mL, then the mass. From that, the purity.

Give it a try, and post the anwer.

I tried on first one. .

Na2CO3 + HCl... > 2NaCl + H2O + CO2

no.of moles (HCl) : (27.92)(0.1131) = 3.157

no.of moles (Na2CO3) : 3.157 / 2 = 1.578

mass (Na2CO3) : (106) (1.578) = 167.26g

I felt that wrong because its bigger than tha sample. .

Indeed. The actual definition of 0.1131 is 0.1131 moles in 1000 mL.

So, in 27.92 mL, you don't have 3.157 moles. (which is already surprising! You have only 0.1131 in 1000 mL, and you got 3.157 moles in only 27.92mL)

Now that it's clearer, try again :)

Note, I noticed that you put 1 mole of HCl in your post. I hope that it's a typo from you, as you did correct it in your calculations.

Na2CO3 + 2HCl... > 2NaCl + H2O + CO2

(0.167 / 0.3215) * 100 = 52 %


The second I did not know how can I write the balancw equation

I just googled soda ash, and it turns out that it is another name for sodium carbonate.

Now, go on :)

So it will be the same equation

Na2CO3 + 2HCl... > 2NaCl + H2O + CO2

no.of moles (HCl) : (4 ml)(0.2) = 8 *10^-4

no.of moles (Na2CO3) : 8 *10^-4/ 2 = 4 *10^-4

mass (Na2CO3) : (106) (4 *10^-4) = 0.0424 g


now how can I get the purity ?

There is a little twist in the question here.

Did you see that from 250 mL of the solution, only 10 mL were used?

Ok, you therefore have 4 x 10^-4 moles of Na2CO3 in 10 mL
Find the number of moles in 250 mL (the concentration remains the same)

From then, find the mass of Na2CO3, to finally obtain the percentage purity.

n = C * V
C= 4*10 ^-4 / 10 = 4*10^-5

n = (4*10^-5) ( 0.25) = 1*10^-5

mass (Na2CO3) : (106) (1*10^-5) = 1.06*10^-3 g

percentage purity = (1.06*10^-3 / 1.5 ) * 100 = 0.07 %

No, you messed up with your calculations.

I don't use the formula, because I never really remember all the formulae I have to learn :o

Ok, my method, proportions:

10 mL -> 4x10^-4 mol
1 mL -> 4x10^-4 / 10 = 4x10^-5 mol
250 mL -> 4x10^-5 * 250 = 0.01 mol

Now continue

In your equation, you did not pay much attention to the units.

n (mol) = C(molmL^{-1}) \times V(mL)

or

n (mol) = C(molL^{-1}) \times V(L)

You found your concentration in mol/mL, and then you took the volume in litres. Either you take both in litres, or both in mL.

10 mL -> 4x10^-4 mol
1 mL -> 4x10^-4 / 10 = 4x10^-5 mol
250 mL -> 4x10^-5 * 250 = 0.01 mol

mass (Na2CO3) : (106) (0.01 ) = 1.06 g

percentage purity = (1.06 / 1.5 ) * 100 = 70.6%


Thanks a lot for your help. .

now really I understood it. .

That's also how most of us learn, right? Doing the problem the wrong way, and right way teaches us where not to go :)

Glad I was of help! :)