Thursday
March 30, 2017

Post a New Question

Posted by on .

Hello, I need help with calculations for lab. I prepared excess NaOH and HCL solutions (unknown conc., ~0.1M). Then I standardized my NaOH solution by titrating known amounts of KHP (Potassium hydrogen phthalate)with it in the presence of phenolphthalein (indicator turns pink at pH 8.3). Then I titrate a known amount of my HCL solution (25.00 mL) with NaOH once again, also in the same indicator. How do I use this information to find out the concentration of both solutions.

In total I have given:
mass KHP and volume NaOH solution required to titrate to pH 8.3.

volume HCl and volume NaOH solution required to titrate to 8.3 pH

I need to find out the concentration of both solutions.

  • Chemistry - ,

    You weighed KHP. Calculate the moles KHP you titrated with NaOH. Moles KHP = grams/molar mass. You should write the equation for KHP + NaOH but it is a 1:1 ratio (1 mole KHP to 1 mole NaOH). Therefore, moles KHP titrated will equal moles NaOH used. Then you know M x L = moles NaOH. You know moles and L, calculate molarity.

    Next you used the standardized NaOH to titrate HCl. Again, the ratio of HCl to NaOH is 1:1. You can use mL acid x M acid = mL base x M base. You know mL acid, mL base and M base. That leaves just one unknown, M acid. Now you know both M NaOH and M HCl.

  • Chemistry - ,

    Are you sure that the fact that it's titrated to 8.2-8.3 pH instead of 7.0 they are still in equal proportions? I was afraid I would have to do pKa's and -Log[conc] here

  • Chemistry - ,

    Yes, I am sure. First, the fact that you are using them for BOTH the standardization AND the titration of HCl, the titration error tends to cancel. However, in the titration of a strong acid (which HCl is) and a strong base (which NaOH is) although the pH at the equivalence point is 7.0, think about the number of mL of extra NaOH needed to make the pH rise from 7.0 to 8.3. The answer is a VERY VERY small amount of NaOH because the titration curve is essentially vertical at that point in the titration. You can calculate it if you wish but it's so close to salt water at that point that one must take into account the Debye-Huckel theory and calculate the activities and use them instead of straight molarity. At any rate, I can tell you that the error is not more than 1 part per thousand and I doubt it is that much. (It's FAR less than 1 drop of NaOH which, on modern burets is about 0.03 mL). But let's just assume its as much as 0.03 mL and you titrated with a total volume of 30 mL. The error is (0.03/30)*100 = 0.1% or 1 part per thousand. I can tell you from experience that it's less than that. Furthermore, you can't read the buret much better than 0.02 mL (and you read it twice). By the way, that's one reason you try to use as much as you can from the titrant buret; i.e., you want that denominator to be large. That is 0.03 mL error out of 10 mL is much much worse than 0.03 mL out of 45 mL (0.3% versus 0.07%). This probably tells you much more than you ever wanted to know about titrations but I hope it's helpful. Put your mind at ease about the indicator error (in this case). (It will affect the results in some titrations but not phenolphthalein in a strong acid/strong base titration.

  • Chemistry - ,

    Thank you very much. Your response was very helpful, thank you so much for taking your time to help people out here

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question