I need to determine how many moles are in my acid-base titration. I used 10 mL o

Question

I need to determine how many moles are in my acid-base titration.
I used 10 mL of HCl in both of my trials.
Initial volume NaOH (mL) Trial 1:
.65 mL
Final volume NaOH (mL) Trial 1:
20.05 mL
Total volume NaOH (mL) used Trial 1:
19.4 mL

Initial volume NaOH (mL) Trial 2:
20.05 mL
Final Volume NaOH (mL) Trial 2:
38.95 mL
Total volume NaOH used (mL) Trial 2:
18.9 mL

Average volume NaOH (mL) used for trial 1&2: 19.15 mL or .01915 L

I believe the balancecd equation for this is: NaOH + HCl -> H2O + NaCl

The molarity of NaOH is 0.1 (I believe)

I need to find the following:

1. Moles of NaOH used in titration

2. Moles of HCl neutralized by NaOH

3. Molarity of unkown HCl

I need to show my work along with the answer, please help. I cannot figure out how to do this.

Explanation / Answer

For volumetric analysis we consider

M1V1 = M2V2 ( millimoles of acid = millimoles of base)

1. Moles of NaOH = Molarity X volume of NaOH = 0.1 X 0.01915 = 0.001915 moles

2.Molarity of HCL = Molarity of NaOH X Volume of NaOH / volume of HCl = 0.1 X 19.15 / 10 = 0.1915 molar

3. moles of HCl = molarity X volume = 0.1915 X 0.01 = 0.001915 moles

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