Tim's Titration Lab

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 w...

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....

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

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 = ?

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 a...

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?

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...

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

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 ?

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

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 appreci...

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

Percentage difference is the same as percentage error?

what answer should I get?

0.00% + -8.06% + -5.82% + 0.00% + -3.42% / 5 = 3.46%

How does he tell me to do this? IM quite confused please let me know