Titration Calculations

[Supreme]

Aye, the worst part of my entire Chemistry A-Level (which I achieved an A in! w00t!)

Could someone please explain to me titration calculations fully? Why they're there? How to work one out? I could master practically everything on the course, and have always thought I can grasp the basics of chemistry, but titrations just lost me. I'm not sure if there's many chemists here, but all help appreciated.

 

[metherion]

Sure thing.

You use titrations to find out the concentration of an unknown solution by measuring it with one you know or can calculate.

First, you need an indicator that will change color yet not interfere with the reaction. With the proper indicator, the color change is the endpoint.

With a titration, you will know two things:
1. The volume of titrant (what's in the buret) that you put in, and
2. Either the concentration or the mass of solute in one of the solutions.

For example, you could titrate HCl with NaOH and (as you do the titration), once you find out the volume of NaOH used to neutralize the HCl, you will find:
a) If you know the mass of HCl you titrated, you can find the molarity of the NaOH.
b) If you know the concentration and volume of HCl, you can find the molarity of the NaOH.
c) If you know the concentration and volume of NaOH, you can figure out the mass of the HCl.

I hope that's making sense, it's easier to explain in person.

The main equations are:

stoichiometric coefficient of titrant *Volume of titrant * concentration of titrant =

stoichiometric coefficient of titrand (what's in the beaker) * volume * concentration.

Metherion

 

[Supreme]

Sure thing.

You use titrations to find out the concentration of an unknown solution by measuring it with one you know or can calculate.

First, you need an indicator that will change color yet not interfere with the reaction. With the proper indicator, the color change is the endpoint.

With a titration, you will know two things:
1. The volume of titrant (what's in the buret) that you put in, and
2. Either the concentration or the mass of solute in one of the solutions.

For example, you could titrate HCl with NaOH and (as you do the titration), once you find out the volume of NaOH used to neutralize the HCl, you will find:
a) If you know the mass of HCl you titrated, you can find the molarity of the NaOH.
b) If you know the concentration and volume of HCl, you can find the molarity of the NaOH.
c) If you know the concentration and volume of NaOH, you can figure out the mass of the HCl.

I hope that's making sense, it's easier to explain in person.

The main equations are:

stoichiometric coefficient of titrant *Volume of titrant * concentration of titrant =

stoichiometric coefficient of titrand (what's in the beaker) * volume * concentration.

Metherion

Thank you for this, could you please show an example?

 

[metherion]

Okay. Sure.

I have .5 L of NaOH of an unknown concentration.

I use 50.00 mL in my buret. (most burets are graded to 1/10 mL, so for sig figs you read to .01 mL).

I get 1 gram of anhydrous KHP (potassium hydrogen phthalate) (Yes, I just TAed this lab on Monday hehe) and add sufficient water to be dissolved. I find that the molecular weight of KHP is 204.23 from the container, so I have (1.000 g * 1 mole/204.23g = ) 4.896 * 10^-3 moles of KHP. I use phenolpthaline as an indicator, which turns pink in basic solutions. Now I add NaOH, and find that it takes me 32.68 mL to turn it just barely pink. (The balances I use in lab give 3 decimal places, and the 5 sig figs on the weight comes from the container of KHP itself. Your lab equipment may give different sig figs.)

So I have NaOH + KHP => NaKP + H2O, so I have 1 mole NaOH per 1 mole KHP. I had 4.896 * 10^-3 moles KHP, so I must have had 4.896 * 10^-3 moles NaOH.

So the concentration of NaOH is 4.896*10^-3 moles / 32.68 mL. Converting to moles per liter, we have 4.896 moles / 32.68 L (10^3 mL/L, 10^3 cancels out 10^-3), which comes to .1498 M NaOH.

Metherion