Topic: Titration Do Now:

Acid-Base Titration A procedure used in order to determine the molarity of an acid or base Known volume of a solution with a known concentration (standard solution) and known volume of unknown concentration and a MAVA = MBVB acid-base indicator needed

Titration Standard solution slowly added to unknown solution As solutions mix: neutralization reaction occurs Eventually: enough standard solution is added to neutralize the unknown solution  Equivalence point

total # moles H+1 ions donated by acid equals Equivalence point total # moles H+1 ions donated by acid equals total # moles H+1 accepted by base total moles H+1 = total moles OH-1

Titration End-point = point at which indicator changes color if indicator chosen correctly: end-point very close to equivalence point

Titration of a strong acid with a strong base 14- Phenolphthalein Color change: 8.2 to 10  pH 7- Equivalence Pt  0-    Volume of 0.100 M NaOH added (ml) 0 ml 40ml 20 ml

MH+1 VH+1 = MOH-1 VOH-1 MH+1 = molarity of H+1 MOH-1 = molarity of OH-1 VH+1 = volume of H+1 VOH-1 = volume of OH-1

Titration Problem #1 In a titration of 40.0 mL of a nitric acid solution, the end point is reached when 35.0mL of 0.100M NaOH is added Calculate the concentration of the nitric acid solution HNO3 + NaOH  H2O + NaNO3

Variables # of H’s = 1 Ma = ? Va = 40.0 mL # of OH’s = 1 Mb = 0.100 M Vb = 35.0 mL

(1)(X) (40.0 mL) = (0.100 M )(35.0mL)(1) X = 0.875 M HNO3

Titration Problem #2 What is the concentration of a hydrochloric acid solution if 50.0 mL of a 0.250M KOH solution is needed to neutralize 20.0mL of the HCl solution of unknown concentration? KOH + HCl  H2O + KCl

Variables # of H’s = 1 Ma = X Va = 20.0 mL # of OH’s = 1 Mb = 0.250 M Vb = 50.0 mL

(1)(X)(20.0 mL) = (0.250 M) (50.0 mL)(1) X = 0.625 M HCl

Titration Problem #3 What is the concentration of a sulfuric acid solution if 50.0mL of a 0.25 M KOH solution is needed to neutralize 20.0mL of the H2SO4 solution of unknown concentration? H2SO4 + 2 KOH  2 H2O + K2SO4

Variables # of H’s = 2 Ma = X Va = 20.0mL # of OH’s = 1 Mb = 0.25M Vb = 50.0mL

(2)(X)(20.0ml) = (0.25M)(50.0ml)(1) X = 0.3125 M H2SO4(sulfuric acid)