1 / 8

4.61 Acid-Base Neutralization

4.61 Acid-Base Neutralization. Hydrochloric acid (75.0 mL of 0.250 M) is added to 225.0 mL of 0.055 M Ba(OH) 2 solution. What is the concentration of the excess H + or OH - ions left in soln? Concepts Balanced Chem Eqn. 4.61 Problem Solving Strategy. # mol acid available

vern
Download Presentation

4.61 Acid-Base Neutralization

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 4.61 Acid-Base Neutralization • Hydrochloric acid (75.0 mL of 0.250 M) is added to 225.0 mL of 0.055 M Ba(OH)2 solution. What is the concentration of the excess H+ or OH- ions left in soln? • Concepts • Balanced Chem Eqn

  2. 4.61 Problem Solving Strategy • # mol acid available • # mol base available • # mol base needed to react with # mol acid available • Find LR and XS • # mol OH- available • # mol OH- that reacts • #mol OH- in excess • Total volume of soln • Molarity of OH-

  3. 4.61: Find LR • Calculate # mol base needed to react with available acid) • (0.075 L x 0.250 M) x (1 mol Ba(OH)2 /2 mol HCl) = 0.009375 mol base needed • Calc # mol base available • (0.225 L x 0.055 M) = 0.012375 mol base available > 0.009375 mol based needed so base = XS and acid = LR

  4. 4.61 Find #mol XS Reagent • Calculate # mol OH-available and #mol OH- that will react with LR = H+. • 0.012375 mol base x 2 mol OH-/ 1 mol base = 0.024750 mol OH-available • 0.009375 mol base x 2 mol OH-/ 1 mol base = 0.018750 mol OH-needed • (0.024750 mol OH- - 0.018750 mol OH-) = excess OH- = 0.00600 mol OH-

  5. 4.61 Find Molarity of XS • Finally calculate molarity of excess OH- • Total volume = 0.300 L • M = #mol OH-/L soln = 0.00600mol/0.300L = 0.0200 M OH-

  6. 4.66 Titration • Standardize a soln of NaOH with a known mass of KHP (monoprotic acid of molar mass = 204.22 g/mol). Ie determine [OH-] of base. • Mass of KHP = 0.1082 g • Volume of base = 34.67 mL • Concepts • Balanced Chem Eqn

  7. 4.66 Problem Solving Strategy • # mol KHP • # mol base needed to neutralize available # mol KHP • Molarity of base

  8. 4.66 Solution • #mol KHP = 0.1082g/204.22g/mol = 5.298E-4 mol KHP • Since neutralization rxn says one mol KHP or # mol H+ reacts with one mol base or one mol OH-, #mol base = 5.298E-4 mol base • #mol base = Mb x Vb • Mb = #mol base/Vb = 5.298E-4 mol/0.03467L = 0.01528 M

More Related