[rycharles]

The experiment was to determine the percent sodium carbonate in the soda ash solid.  There are 2 equilibrium points for the experiment. 

grams of Na2CO3 = 0.2018g (dissolved in 50ml of water)
1st end point volume: 18.78 ml
2nd end point volume: 18.55 ml

Na2CO3(s) + 2HCl (aq) --> CO2(g) + H2O(l)

Can anyone give a hint in determining the molarity of the HCl?

Is this the right idea?

grams of Na2CO3 --> moles of Na2CO3 --> moles of HCl --> Molarity of HCl

Also, what volume of solution do I use since I have two end points?  Do I use the first one or the second one or combine the volume?

[discodermolide]

The experiment was to determine the percent sodium carbonate in the soda ash solid.  There are 2 equilibrium points for the experiment. 

grams of Na2CO3 = 0.2018g (dissolved in 50ml of water)
1st end point volume: 18.78 ml
2nd end point volume: 18.55 ml

Na2CO3(s) + 2HCl (aq) --> CO2(g) + H2O(l)

Can anyone give a hint in determining the molarity of the HCl?

Is this the right idea?

grams of Na2CO3 --> moles of Na2CO3 --> moles of HCl --> Molarity of HCl

Also, what volume of solution do I use since I have two end points?  Do I use the first one or the second one or combine the volume?

Try drawing the balanced equations out properly, in your equation where has the Na and the Cl disappeared to?

[rycharles]

I forgot the 2 NaCl.  Here is the balanced equation.

Na2CO3(s) + 2HCl (aq) --> CO2(g) + H2O(l) + 2NaCl (aq)

Can I use the M1V1 = M2V2 equation here?

M(HCl)*V(HCl) = M(Na2CO3)*V(Na2CO3)

M(HCl) = [M(Na2CO3)*V(Na2CO3)]/V(HCl)

M(Na2CO3) --> 0.2018gNa2CO3(mol/105.99g) = 0.0019040 mol Na2CO3

V(Na2CO3) --> 50.00 ml

V(Na2CO3) --> What volume do I use? 1st end point, 2nd endpoint, or combine them?

[discodermolide]

I forgot the 2 NaCl.  Here is the balanced equation.

Na2CO3(s) + 2HCl (aq) --> CO2(g) + H2O(l) + 2NaCl (aq)

Can I use the M1V1 = M2V2 equation here?

M(HCl)*V(HCl) = M(Na2CO3)*V(Na2CO3)

M(HCl) = [M(Na2CO3)*V(Na2CO3)]/V(HCl)

M(Na2CO3) --> 0.2018gNa2CO3(mol/105.99g) = 0.0019040 mol Na2CO3

V(Na2CO3) --> 50.00 ml

V(Na2CO3) --> What volume do I use? 1st end point, 2nd endpoint, or combine them?

Why are there two end points?

[rycharles]

I forgot the 2 NaCl.  Here is the balanced equation.

Na2CO3(s) + 2HCl (aq) --> CO2(g) + H2O(l) + 2NaCl (aq)

Can I use the M1V1 = M2V2 equation here?

M(HCl)*V(HCl) = M(Na2CO3)*V(Na2CO3)

M(HCl) = [M(Na2CO3)*V(Na2CO3)]/V(HCl)

M(Na2CO3) --> 0.2018gNa2CO3(mol/105.99g) = 0.0019040 mol Na2CO3

V(Na2CO3) --> 50.00 ml

V(Na2CO3) --> What volume do I use? 1st end point, 2nd endpoint, or combine them?

Why are there two end points?

There are two end points because

CO3^2- + H⁺ --> HCO3^-  (first equivalence point) phenolpthalein indicator - from pink to clear color
HCO3^- + H⁺ --> H2CO3  (second equivalence point) methyl orange indicator - from orange to pink color

[discodermolide]

I forgot the 2 NaCl.  Here is the balanced equation.

Na2CO3(s) + 2HCl (aq) --> CO2(g) + H2O(l) + 2NaCl (aq)

Can I use the M1V1 = M2V2 equation here?

M(HCl)*V(HCl) = M(Na2CO3)*V(Na2CO3)

M(HCl) = [M(Na2CO3)*V(Na2CO3)]/V(HCl)

M(Na2CO3) --> 0.2018gNa2CO3(mol/105.99g) = 0.0019040 mol Na2CO3

V(Na2CO3) --> 50.00 ml

V(Na2CO3) --> What volume do I use? 1st end point, 2nd endpoint, or combine them?

Why are there two end points?

There are two end points because

CO3^2- + H⁺ --> HCO3^-  (first equivalence point) phenolpthalein indicator - from pink to clear color
HCO3^- + H⁺ --> H2CO3  (second equivalence point) methyl orange indicator - from orange to pink color

So you should combine them.

[Borek]

General calculation of titration results:

http://www.titrations.info/titration-calculation

But there are things that I don't like in your description. How come the volume of the second end point is lower than the volume of the first end point? During titration of the sodium carbonate you can observe two endpoints, but the volume to reach the second should be exactly twice the volume of the titrant necessary to reach the first, unless you are titrating not just a sodium carbonate, but some combination of sodium carbonate and other compounds. Compare discussion of the Warder titration.

[rycharles]

The total volume used to reach the 2nd end point was

18.78ml + 18.52ml = 37.30ml (My question is, can I use this value to determine the molarity of HCl?)

This is what I did.

calculating moles of Na2CO3

0.2018g Na2CO3*(1mol/105.99g) = 0.0019040 mol Na2CO3

since the mole ratio is 1:2 for Na2CO3 and HCl (according to the equation)

0.0019040mol Na2CO3*(2mol HCl/1mol Na2CO3) = 0.0038079 mol HCl

so now I have the moles of HCl and I divide by the total volume of HCl (18.78+18.52=37.30) to get the molarity

37.30 ml*(L/10³ ml) = 0.03730 L

0.0038079 mol HCl/0.03730 L = 0.10209 M HCl

Can anyone tell me if I made a mistake or if I am on the right path?

[Borek]

The total volume used to reach the 2nd end point was

18.78ml + 18.52ml = 37.30ml (My question is, can I use this value to determine the molarity of HCl?)

Ah, so these were separate volumes, not the total burette reading. Makes sense now.

0.10209 M HCl

Looks OK to me, although I would wrote 0.1021M

[rycharles]

Now that I have the molarity of my HCl, how do I get the percent of my sodium carbonate and sodium oxide from my original soda ash.

This is what I did.

molarity of my standard HCl = 0.1020₉ M
weight of my unknown sample: 0.2015g
1st end point: 3.89mL
2nd end point: 3.99mL
total volume: 3.89 + 3.99 = 7.88 mL

7.88ml*(1L/10³mL) = 0.00788 L HCl

0.00788L*(0.10209mol/L) = 0.00080447 mol HCl

0.00080447 mol HCl*(1mol Na2CO3/2mol HCl) = 0.00040224 mol Na2CO3

0.00040224 mol Na2CO3*(105.99 g Na2CO3/1mol Na2CO3) = 0.042633 g Na2CO3

weight percent of Na2CO3 from unknown soda ash sample --> (0.042633 g Na2CO3 / 0.2015 g sample)(100%) = 21.58%

Does this calculation look correct or are there any mistakes?  Also, how do I calculate the sodium oxide (Na2O) from my unknown?

[Borek]

Looks OK.

How many moles of Na₂O per one mole of Na₂CO₃?

[rycharles]

Na2CO3(aq)  Na2O(aq) + CO2(g)

Is this right? If it is then the mole to mole ratio is 1:1.

I can get the value of Na2O by comparing the moles of Na2CO3 and Na2O.

so,

moles of 0.00040224 Na2CO3 (calculated earlier)

0.00040224 mol Na2CO3*(1mol Na2O/1mol Na2CO3) = 0.00040224 mol Na2O

Can I use this value to calculate the Na2O percent from my unknown sample?

0.00040224 mol Na2O*(61.98g Na2O/1mol Na2O) = 0.024931g Na2O

(0.024931g Na2O/0.2015g sample)*(100%) = 12.37%

Does this look right?

[Borek]

Yes.

[rycharles]

Greatly appreciated!

[rycharles]

Can anyone give any details about how to do a gran plot for this experiment?  The formula given was:

V_a10^+pH=((1*γ_B)/(K_a*γ_BH⁺))