Lab Experiment 350
Standardizing a Sodium Hydroxide Solution

 

I won't say that titrations are tedious or anything, but there are certainly more exciting lab experiments out there. Certainly, that time I accidentally held a piece of magnesium over my bunsen burner (it was an accident, I swear!) and ignited it into a glowing ball of fury was much more exciting than titrating one dumb fluid into another. Oh, it turned colors? Hurray.

 

(silence)

 

But we have to do it, because as my previous lab instructor said, "If you aren't good at it, babies will die." I have no idea what he means. I vaguely recall it having something to do with titration being necessary for prenatal care, but I'm hardly interested in babies so I hardly paid it any attention.

 

On another note, if you do "accidentally" ever ignite a piece of magnesium with your bunsen burner, do not every try and extinguish the fire with water. Water only feeds the reaction.......

 

So next up in our titration series is the standardization of NaOH. Because we're already familiar with titration terms and all that malarchy, I won't reiterate it all again here. There are, however, a few important points that are applicable to NaOH that weren't mentioned in our previous endeavors....

 

1. Strong base solutions, such as our NaOH in this module, must be protected from CO2 in the air.

I read this in my lab manual and was disappointed because once again, education feeds us a lot of information and doesn't give us the why. So why must NaOH be protected from the air? Well, I ran into a very interesting and informative article when I was researching the answer to this question about how a certain company plans to purify the air released from factories of excess CO2 by running it through a tower to react with NaOH. NaOH is hygroscopic and absorbs the CO2 from the air, remember our ionic hydrate? Kind of the same, except it's absorbing CO2 instead of H2O. We don't want this to happen because we want our NaOH to be as pure as possible. Hence, we keep it away from the CO2 infested air.

2. Sodium hydroxide gradually etches glass. Therefore, NaOH should not be stored in glass containers.

 

In this experiment, dilute HCl was our titrant and phenolphthalein is the indicator.

 

About the indicator: phenolphthalein is colorless in acidic solution and red in basic solution. As it passes from these conditions, it will appear pale pink. Pale pink if the end point for this titration as it indicates neutralization.

 

So let's get to the calculations, since they're the last thing here that's really different than the previous titration I covered.

 

The chemical equation given is H3O+(aq) + OH-(aq) <<<---------------->>>2H2O(l)

The given stoichiometry indicates a 1:1 ratio of NaOH to HCl and therefore:

the number of moles of NaOH = the number of moles of HCl

 

So we already know from the last lab that (MHCl)(VHCl)= number of moles of HCl

And looking at the statement above we see that number of moles of NaOH = (MNaOH)(VNaOH)

So from that we can derive the equation (MHCl)(VHCl)=(MNaOH)(VNaOH)

 

Remember M is in mol L^-1

and V is volume in LITERS, always liters, L

 

Then, the equation is just rearranged depending on the values given and the value that needs to be determined. Let's play around with it..........

 

From my book: a 25.00mL sample of 0.1030M HCl is titrated with exactly 24.42mL of dilute NaOH solution. What is the concentration of the NaOH solution?

 

General breakdown of values we have to work with:

We have 25.00mL of HCl
The molarity of that HCl is 0.1030M

We also have 24.42mL of NaOH solution

The molarity of the NaOH is unknown, that's what we need to find!

 

Well, once again, let's start with doing our conversions just to make sure everything is all proper and ready to be plugged into our equation.

25.00mL x 1L/1000mL = 2.500 x 10^-2 L for our HCL

and 24.42mL x 1L/1000mL = 2.442 x 10^-2 L for our NaOH solution

Now we can plug them all into our equation (m1)(v1)=(m2)(v2)

 

(0.1030M)(2.500x10^-2L)=(MNaOH)(2.442x10^-2L)

 

Because we need to find the molarity of NaOH, we rearrange the equation to look like this:

 

MNaOH= (0.1030M)(2.500x10^-2L)/2.442x10^-2L

 

Okay! We are ready to solve!

Note when solving that the like terms will cancel out.

Top first! 0.1030(2.500) = 0.2575       (no 10^-2 necessary, because they cancelled out)

Divide that by 2.442 = 0.1054M           (again, no 10^-2 necessary)

Note that our answer is in terms of M, when we were asked to give it in mol L^-1, they are the same thing, so just change it.

 

Soooo, our answer is: 0.1054 mol L^-1 is the concentration of NaOH