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Tim's Titration Lab
Questions (6)
If 10.0 mL of 0.250 mol/L NaOH(aq) is added to 30.0 mL of 0.17 mol/L HOCN(aq), what is the pH of the resulting solution?
1 answer
1,931 views
Lab: Determining Ka of Acetic Acid
Purpose: The purpose of this experiment is to determine the molar concentration of a sample of
1 answer
717 views
How would I calculate this average, given the minus signs in the calculation?
Average = 0.00% + -8.06% + -5.82% + 0.00% + -3.42%
11 answers
422 views
[(exp value - accepted value)/accepted value]*100 = percent error.
For exp value, do I have to average my five trials?
5 answers
610 views
Trial 1
Volume of HCL is 2.51 mL Molarity of the HCL is 0.10 M Mols HCL = L x M = 0.0251 L x 0.10 M = 0.00251 mol/L The equation
7 answers
711 views
Purpose: The purpose of this experiment is to determine the Ksp for calcium hydroxide.
Materials are: *50.0 mL of 0.10 M HCl(aq)
9 answers
1,225 views
Answers (12)
How does he tell me to do this? IM quite confused please let me know
0.00% + -8.06% + -5.82% + 0.00% + -3.42% / 5 = 3.46%
what answer should I get?
Percentage difference is the same as percentage error?
[(exp value - accepted value)/accepted value]*100 = percent error. For exp value, do I have to average my five trials?
Beleive me, I will this is truly a great site with great experts!
My friend, you are amazing! DrBob for Chem President!
Oh really sorry it was my own mistake:P After I repeat this same technique for the next four trials, another question asks me to conduct a search to determine the actual value for the solubility product of calcium hydroxide(which you did, I really
Im going to try to solve for the first trial please let me know if I am correct. I will send it within the next 10 minutes
I see, so for the first trial my volume would be 2.51 ml For the second trial, the volume would be 2.44 ml For the third trial, the volume will be 2.45 mL For the fourth trial, the volume will be 2.50 mL For the fifth trial, the volume will be 2.49 mL ?
To get volume of HCL, would I do this? 2.51+2.44+2.45+2.50+2.49 / 5 = 2.48 ?
Here are my results: *Initial volume of HCL(mL) Trial 1- 0.0 Trial 2- 2.51 Trial 3- 4.95 Trial 4- 7.4 Trial 5- 9.9 *Final volume of HCL (mL) Trial 1- 2.51 Trial 2- 4.95 Trial 3- 7.4 Trial 4- 9.9 Trial 5- 12.39 *Volume of HCL added (mL) Trial 1- 2.51 Trial