Lab Experiment #349
Standardizing a Hydrochloric Acid Solution
Standard Solution: a solution of preciesely known concentration.
Standardizing: the process of determining the concentration of a solution.
Titration: the careful addition of one solution from a buret to another substance in a flask until all the substance in the flask has reacted.
Titrant: the solution added from the buret.
Indicators: an organic dye added to flask substance to indicate acidic or basic conditions.
End Point: the color change of the indicator that occurs upon addition of only a small amount of titrant.
If one of the reacting substances in a titration has been standardized, then data from the titration can be used to calculate the concentration of the other substance.
One of the reactants must be a primary standard.
Primary standards must follow these rigorous guidelines:
1. available in very pure form
2. resonably soluble
3. stable in the pure form in the solution
4. nonhygroscopic and easily dried
5. a compound with a molecular mass above 100
The commercially available concentrated HCl is approximately 12M, but its concentration isn't sufficiently constant to permit its use for calculating the concentration of the dilute solution directly. For this reason, a solution of approximately the desired concentration of HCl is prepared and then standardized with a primary standard, such as Na2CO3 (sodium carbonate).
In titrating Na2CO3 with HCl, the diluted HCl is considered the titrant because it is being dispensed from the buret into an Erlenmeyer flask of distilled water, Na2CO3, and the indicator (in this module, bromcresol green).
About the indicator:
bromcresol green is yellow in acid solution and blue in base solution. In HCl titration, the solution absorbs a significant amount of CO2 produced by the titration reaction below:
2H3O+(aq)+CO3-2(aq) <<<--------[H2CO3]-------->>>3H2O(l) + CO2(g)
<<<----->>> indicates reaction is occuring back and forth on both sides
The CO2 produced and absorbed by the solution is in equillibrium with the H2CO3 (carbonic acid), and consequently both the dilute HCl solution we are titrating and the carbonic acid contribute to the acidity of the the solution. This may produce a premature endpoint. To remove excess CO2, the titration is interrupted just as the indicator begins to flash and boiled until the solution changes bad to the color it had been before the indicator began to flash. It is then cooled down to room temperature and titration is continued.
From the above chemical equation, we are indicated that two moles of HCl are required to react with one mole of Na2CO3, the ratio is 2:1.
The following describes the process of determining concentration of dilute HCl solution:
The number of moles of HCl = 2(number of moles of Na2CO3) -because we have a 2:1 ration based on our equation.
We can further build our equation by:
Number of moles of HCl = (MHCl)(VHCl)
whereas: MCl = molarity of HCl in mol^-1 and VHCl = volume of HCl in L
Number of moles Na2CO3= mass of Na2CO3, g/106.0 g mol^-1
Whereas the weight of our Na2CO3 in grams divided by the molar mass of Na2CO3 gives us the number of moles of Na2CO3 we had in our solution.
Now, here's the FUN. FUN. FUN. We know that number of moles of HCl is equal to 2 multiplied by the
number of moles of Na2CO3 so....
(MHCl)(VHCl)=2(mass of Na2CO3, g)/106.0g mol^-1
An important note here is that (VHCl) must be converted to liters before being plugged into the equation!! Here's a recap on how to convert the mL of HCl into L so that everything can be super happy good.....
#mL x 1L/1000mL
so basically you just divide your value in mL by 1000 to convert, the mL cancel out and we're left with lovely, wonderful liters.
So let's play pretend that we have, and don't think me brilliant because I'm stealing this example from my handy dandy laboratory manual and just re-wording it, a 0.1338g sample of solid Na2CO3 which is dissolved in some arbitrary amount of water and titrated with exactly 24.63mL of dilute HCl solution. What is the concentration of the HCl solution?
First, let's break down this problem to determine exactly which values we are being given. Analyzing anything before tackling it makes your attack more organized and efficient ;)
We have
0.1338g Na2CO3
24.63 mL HCl
Doesn't seem like much, but actually, it's perfect! That's all we need!
Next, let's convert the mL of HCl into L to get that out of the way.
We have 24.63mL of dilute HCl solution so 24.63mL x 1L/1000mL
24.63X1= 24.63, and 24.63/1000 = 2.463 x 10^-2L
So now that we have our liters of HCl, we can plug everything into our fun equation!
(MHCl)(2.463x10^-2 L HCl) = 2(0.1338g Na2CO3)/106.0g mol^-1
But that's not quite proper, is it? We have to have the Molarity of HCl isolated on one side of the equation to determine a result! A little rearranging is in order, I think....
(MHCl)= 2(0.1338g Na2CO3)/(106.0g mol^-1)(2.463x10^-2)
All that's left to do is solve this bad boy and we're free to go back to......... well, much more of this, quite honestly....... let's solve it anyway though, shall we?
Top first! 0.1338x2 = 0.2676
Now bottom! (106.0)(0.02463)= 2.610
Now divide! 0.2676/2.610 = 0.1025 mol L^-1
so our answer is 0.1025 mol L^-1 is the concentration of the dilute HCl solution
***So I was just thinking and maybe you're wondering where I got the molar mass for Na2CO3. Well, it was given beforehand, but not in the particular problem I broke down, making the problem seem a bit incomplete. I think I'll do a page specifically on calculating molar mass over the weekend or maybe next week sometime. This week is my laboratory midterm and so we're all labs, always.***