Acid-base titration is a very important topic that is taught in general chemistr

Question

Acid-base titration is a very important topic that is taught in general chemistry however it requires an understanding of the chemistry involved and an ease with calculations pertaining to concentration changes in order to determine the pH of solutions. The plot of the pH of a strong acid solution versus the volume of strong base is an example of a titration curve. In this extra credit assignment you are asked to calculate the pH for a 10.0 mL sample of 0.10 M HCl titrated in 1.00 mL increments with 0.10 M NaoH. Co .00 Below is a table for you to record the pH of the HCl solution throughout the titration. On the following pages are two approaches to determine the pH; the first method is typically taught and the second method is an alternative At the end you will be asked to answer a few survey questions to provide feedback about your impressions of the two methods. Your feedback will not be personally identifiable but will be extremely informative for we who teach acid-base titration. Please read through both methods before picking which one to use. vol Base (mL) Tot. Vol (mL) Moles of H. Moles of oH Difference [H"toHl (M) pH 10

Explanation / Answer

Ans. Given, Molarity of HCl = 0.10 M

            Volume of HCl = 10.0 mL = 0.010 L

Moles of HCl in given volume = Molarity x Volume in liters

                                    = 0.10 M x 0.010 L

                                    = (0.10 mol/ L) x 0.010 l                                              [1 M = 1 mol/L]

                                    = 0.001 mol

Balanced reaction:      HCl(aq) + NaOH(aq) --------> Na+Cl-(aq) + H2O(l)

Stoichiometry: 1 mol HCl neutralized 1 mol NaOH.

#1. Tube 1 – Base volume (of NaOH) = 0 mL

            Volume of HCl (in each tube) = 10.0 mL

Total volume = 10.0 mL

[HCl] = 0.10 M            - same as original since no dilution takes place.

Now,

pH = -log [H+] = -log (0.10) = 1.000

            Note: 1 mol HCl donates 1 mol H+.

#2. Tube 2 – Base volume (of NaOH) = 1.0 mL (= 0.001 L)

Step 1: Moles of NaOH in 1.0 mL = 0.10 M x 0.001 L = 0.0001 mol

            Moles of OH- = moles of NaOH = 0.0001 mol

Volume of HCl (in each tube) = 10.0 mL

            Moles of H+ = Moles of HCl = 0.001 mol

Step 2: Moles of H+ in excess = Initial moles of H+ - Moles of H+ neutralized by OH-

                                    = 0.001 mol – 0.0001 mol

                                    = 0.0009 mol

Therefore, after neutralization, there would be 0.0009 mol H+ left behind in the tube.

Step 3: Total volume = 11.0 mL = 0.011 L

Molarity of H+ after neutralization = Moles of H+ in excess / Volume of reaction mixture in L

                                                = 0.0009 mol/ 0.011 L

                                                = 0.08182 mol/ L

                                                = 0.08182 M

Therefore, [H+] after neutralization = 0.08182 M

Step 3: pH = -log [H+] = -log (0.08182) = 1.087

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