Titration Troubles: Where Did We Go Wrong?

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SUMMARY

The forum discussion centers on a titration lab involving sodium hydroxide (NaOH) and potassium hydrogen phthalate (KHP). The participants calculated that approximately 199.7 mL of NaOH is required to titrate 0.4 g of KHP, using the equation C1V1=C2V2 for concentration and volume calculations. A significant issue identified was the use of a concentrated NaOH solution instead of the intended dilute solution, which would prevent a color change in phenolphthalein. Accurate weighing of KHP and careful titration techniques are emphasized as critical for successful titration outcomes.

PREREQUISITES
  • Understanding of titration principles and procedures
  • Familiarity with molarity calculations and the equation C1V1=C2V2
  • Knowledge of the chemical properties of NaOH and KHP
  • Experience with laboratory equipment such as burettes and analytical balances
NEXT STEPS
  • Learn about accurate weighing techniques for solid reagents in titrations
  • Study the preparation of standard solutions and their importance in titrations
  • Research common errors in titration procedures and how to avoid them
  • Explore the use of phenolphthalein as an indicator in acid-base titrations
USEFUL FOR

Chemistry students, laboratory technicians, and educators involved in teaching or conducting titration experiments will benefit from this discussion.

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Homework Statement



We had a titration lab where we took 1.0 g of NaOH and added it to 50 mL of water. We then took that solution and added 10 mL of it to 500 mL of water, producing a 510 mL solution in total (this became the titrant). We then took 0.4 g of potassium hydrogen phthalate and added it to 50 mL of water (this was the analyte).

Our group filled the burette 4 times and didn't get any colour change in the phenolphthalein. Where did we go wrong?

Homework Equations



C1V1=C2V2 and unit conversions

The Attempt at a Solution


[/B]
I calculated that you need 199.7 mL of NaOH to titrate the KHP; can someone please tell me whether I'm right or not?

This is how I calculated it
(0.4 g KHP) * (1 mol KHP / 204.22 g KHP) = 0.019587 mol KHP and since it's monoprotic also 0.0019587 mol NaOH

next the NaOH

1 g NaOH * (1 mol NaOH / 39.997 g NaOH) = 0.025002 mol NaOH

Now using C1V1=C2V2,

(0025002 mol NaOH)/(50 mL*1L/1000mL) <--- this is the initial concentration, C1
10 mL * 1L / 1000 mL <----- this is the initial volume, V1
the final volume is 510 mL NaOH or 0.510 L
rearranging and solving, I got 0.009805 M for the concentration

C = n/V
V = n/C
V = (0.0019587 mol NaOH)/(0.009805 mol/L NaOH) = 0.1997 L or 199.7 L

Any help would be appreciated. Thanks!
 
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From first hand experience, I found that a student took the sample to be titrated from a reagent of concentrated sample, rather than the dilute sample meant for titration. When the teacher found out what bottle they took the samples from , the teacher said he could have been titrating all day before the color would change. The base was too concentrated.
 
Zayn said:
I calculated that you need 199.7 mL of NaOH to titrate the KHP; can someone please tell me whether I'm right or not?

Around 200 mL is what I got.

(0.4 g KHP) * (1 mol KHP / 204.22 g KHP) = 0.019587 mol KHP and since it's monoprotic also 0.0019587 mol NaOH

Either 0.019587 or 0.0019587, but I bet it is just a typo.

next the NaOH

1 g NaOH * (1 mol NaOH / 39.997 g NaOH) = 0.025002 mol NaOH

Now using C1V1=C2V2,

(0025002 mol NaOH)/(50 mL*1L/1000mL) <--- this is the initial concentration, C1
10 mL * 1L / 1000 mL <----- this is the initial volume, V1
the final volume is 510 mL NaOH or 0.510 L
rearranging and solving, I got 0.009805 M for the concentration

Much faster is to realize your 510 mL of the solution contains 10/50*1 g of NaOH.

You never said what the burette volume was, they come in many sizes.

Plenty of things that could go wrong, but planning a titration that requires adding titrant to burette is in general a faulty approach. Shouldn't you be titrating 10 mL of the KHP solution?
 
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I only read this very quickly and not in great detail.

The KHP is a STANDARD which is used for finding what is the concentration of the prepared NaOH titrant. The titrant was not prepared accurately but you must weigh the KHP accurately, even if you take an aloquot of its solution to use for the titration. Take care of all calculations in advance.
 
I took one more but still fast look at the original post. Yes, pay careful attention to all needed calculations ahead of time, and also check your computations and calculations to know and predict the effect of sample size between the KHP and amount of NaOH titrant.
 
This part of the procedure,
We had a titration lab where we took 1.0 g of NaOH and added it to 50 mL of water. We then took that solution and added 10 mL of it to 500 mL of water, producing a 510 mL solution in total (this became the titrant).
if you handle the calculation correctly will predict a CONCENTRATION of 0.009803 M of NaOH titrant; and you would now have 510 ml. of this titrant.

Now when you try to use KHP to check the actual concentration of NaOH titrant, you WILL NOT want to take all of the 510 ml. of this titrant. You would take smoe much smaller aloquot. (Formula weight of NaOH, 39.99 grams per mole. Formula weight of Potassium BiPhthalate, ?--------okay; KHP is 204.22 grams per mole.)

Watch the quantity of KHP weighed very carefully and accurately. You did not weight exactly 0.4 grams KHP. You would or should weight to the nearest 0.0001 gram or better. If you were to weigh exactly 0.4000 grams of KHP, how many MOLES is this?
You will then need to decide how much of your titrant you need to titrate this 0.4000 grams of KHP.
 
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