 # 10.2 Neutralization and Acid-Base Titrations Learning Goal … …use Stoichiometry to calculate volumes and concentrations in a neutralization reaction …

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10.2 Neutralization and Acid-Base Titrations Learning Goal … …use Stoichiometry to calculate volumes and concentrations in a neutralization reaction … determine the pH of a solution with an excess reactant

recall: Acid + Base ---> Water + salt this is known as neutralization a SALT is an ionic compound that is composed of the anion from an acid and a cation from a base HNO 3 (aq) + NaOH (aq)  H 2 O (l) +NaNO 3 (aq) anion cation acid+ base  water + salt if we mix equal amounts of the acid and the base, all acidic and basic properties will be neutralized and the pH of the mixture will be 7 to neutralize an acid or a base we can thus add equal number of moles of the opposite substance

What volume of 0.250 mol/L sulfuric acid is needed to react completely with 37.2 mL of 0.650 mol/L potassium hydroxide? H 2 SO 4 (aq) +2 KOH (aq)  K 2 SO 4 (aq) + 2 H 2 O (l) nCVnCV 0.250 mol/L 0.0372 L 0.650 mol/L 0.02418 mol0.01209 mol 0.0484 L  48.4 mL of sulfuric acid is needed to completely react with the KOH.

25.0 mL of 1.0 M hydrochloric acid reacts with 25.0 mL of a 0.50 mol/L solution of NaOH (aq). How many moles of excess reactant remain? What is the concentration of excess reactant? What is the pH of the final solution? NaOH (aq) +HCl (aq)  NaCl (aq) +H 2 O ( l ) nCVnCV 0.50 mol/L 0.0250 L 1.0 mol/L 0.025 mol0.0125 mol 0.0250 L i u xs 0.0125 mol 0.025 mol0.0125 mol 0 [HCl] = n / V = 0.0125 mol / 0.050 L = 0.25 mol/L pH = - log (0.25) = 0.60  0.0125 mol of excess HCl remains, the [HCl] is 0.25 mol/L and the pH of the final solution is 0.60.

HOMEWORK p466 #1-5 Read p466-469 “Titrations” p470 #1-9 CAN I … …use Stoichiometry to calculate volumes and concentrations in a neutralization reaction

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