Question

Strong-Strong Titrations

Experiment 1: 45.0 mL of 0.125 M HBr is being titrated with 0.375 M NaOH.

C. Point C: 15.0 mL of the titrant has been added.

                                    1. How many moles of HBr are in your initial sample?

                                    2. How many moles of H₃O⁺are in the initial sample?   

                                    3. How many moles of NaOH have been added?   

                                    4. How many moles of OH^-have been added?

                                    5. How many moles of H₂O have been formed from neutralization?

                                                      (HINT:Is H₃O⁺or OH^-the limiting reactant?)

                                    6. What is the pH of the solution now?   

                                    7. This point in the titration is called the:                                   the __________________ point

D. Point D: 25.0 mL of the titrant has been added.

                                    1. How many moles of HBr are in your initial sample?

                                    2. How many moles of H₃O⁺are in the initial sample?

                                    3. How many moles of NaOH have been added?   

                                    4. How many moles of OH^-have been added?

                                    5. How many moles of H₂O have been formed from neutralization?   

                                                      (HINT:Is H₃O⁺or OH^-the limiting reactant?)

                                    6. How many moles of OH^-are left over?

                                    7. What is the total volume of the solution after the 25.0 mL addition?   

                                    8. What is the concentration of the OH^-in the solution now?

                                    9. What is the pOH of the solution now?   

                                    10. What is the pH of the solution now?                   

Answer 1

HBr(09.) + NaOH(aq.) - NaBr + H₂O 2341 0-125M 16 17 moles 10 ili 12 13 14 15 16 17 Point - c. 45.0mL 0.375 M Initial No. of moles of HBr in sample = 450 0.125 -0.005625 HBr is a strong acid thes No. of moles of HBr = No. of moles of 1200 = 0.005625 moles (3 No. of moles of NaoH added = MV ( = 0.375 x 15 1000 No. of moles of one = No. of moles of NaoH = 0.005 625 moles = 0.005625 moles No of moles of oHo = No of moles of How = 0.005625 Thus LIR. is not present. hot oto yahoo No. of moles of Ho formed = 0.005625 6 pH = 7.0 Cat 25°C) solution will be Neatral. & Equivalent point. Point - 0 0.005 625 moles Ⓡ 0.005625 moles Vome) - 0.315 × 25 = 0.009375 moles Ⓡ = MV(ML) = © = 4 = 0.009375 males ③ LOR. = HBr Thus No. of moles of the formed = 0·005625 © L miting Reagent = 406 No. of moles of ohe Left = 0.00935-0.005695 = 0·003750 males 6 Total volume of solution - 45+25 = 70.0mL @ron] = 0.00375 x1000 - 0105357M - 51354x102 molt polt=-109 (5-35 7x10-2) = 3-109 (5:357) = 2-0.7289 = 1.271 18 19 20 21 1000 22 23 24 25 26 27 29 70 30 pH = 14-port - 19.7989