Saturday
April 30, 2016

# Posts by Tim's Titration Lab

Total # Posts: 18

chemistry
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?
April 5, 2008

chemistry
Lab: Determining Ka of Acetic Acid Purpose: The purpose of this experiment is to determine the molar concentration of a sample of acetic acid and to calculate its Ka. Materials: 25 mL pipet and bulb burette 2x150 mL beaker 125 mL Erlenmeyer flask acetic acid solution sodium ...
April 5, 2008

chemistry
How does he tell me to do this? IM quite confused please let me know
April 4, 2008

chemistry
0.00% + -8.06% + -5.82% + 0.00% + -3.42% / 5 = 3.46%
April 4, 2008

chemistry
April 4, 2008

chemistry
How would I calculate this average, given the minus signs in the calculation? Average = 0.00% + -8.06% + -5.82% + 0.00% + -3.42% / 5 = ?
April 4, 2008

chemistry
Percentage difference is the same as percentage error?
April 4, 2008

chemistry
[(exp value - accepted value)/accepted value]*100 = percent error. For exp value, do I have to average my five trials?
April 4, 2008

chemistry
[(exp value - accepted value)/accepted value]*100 = percent error. For exp value, do I have to average my five trials?
April 4, 2008

chemistry
Beleive me, I will this is truly a great site with great experts!
April 4, 2008

chemistry
My friend, you are amazing! DrBob for Chem President!
April 4, 2008

chemistry
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 appreciate that). Compare ...
April 4, 2008

chemistry
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 for the titration is Ca()H)2 + 2HCL --> CaCl2 + 2H2O I will now take 0.5 mols HCl to give mols of OH- 0.00251 mol/L x 0.5 = 0.001255 mol/L mols ...
April 4, 2008

chemistry
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
April 4, 2008

chemistry
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 ?
April 4, 2008

chemistry
To get volume of HCL, would I do this? 2.51+2.44+2.45+2.50+2.49 / 5 = 2.48 ?
April 4, 2008

chemistry
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 2- 2.44 Trial 3- 2.45...
April 4, 2008

chemistry
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) *bromophenol blue indicator *50.0 mLsaturated calcium hydroxide *50.0 mL buret *125 mL Erlynmyer Flask *50.0 mL of distilled water *10.0 mL pipet ...
April 4, 2008

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