Presentation is loading. Please wait.

# Thursday, May 15 th : “A” Day Friday, May 16 th : “B” Day Agenda  Collect Crossword Puzzle  Collect Concept Review  Ch. 15 Review – pass back/warm-up.

## Presentation on theme: "Thursday, May 15 th : “A” Day Friday, May 16 th : “B” Day Agenda  Collect Crossword Puzzle  Collect Concept Review  Ch. 15 Review – pass back/warm-up."— Presentation transcript:

Thursday, May 15 th : “A” Day Friday, May 16 th : “B” Day Agenda  Collect Crossword Puzzle  Collect Concept Review  Ch. 15 Review – pass back/warm-up  Ch. 15 Test Looking Ahead Mon/Tues: Finals Review Wed/Thurs: Fun Lab/Sr. Finals (Thurs) Fri/Tues: Sr. Finals (Fri)/2 nd Semester Final Jeopardy

Ch. 15 Review/Warm-up #21 Write an equation for the reaction between hydrocyanic acid, HCN, and water. Label the acid, base, conjugate acid and conjugate base. HCN + H 2 O ? HCN + H 2 O CN - + H 3 O + Acid Base Conjugate Base Conjugate Acid

Ch. 15 Review/Warm-up #40 If the hydronium ion concentration of a solution is 1.63 X 10 -8 M, what is the hydroxide ion concentration? [H 3 O + ] [OH - ] = 1.00 X 10 -14 M (1.63 X 10 -8 M) [OH - ] = 1.00 X 10 -14 M [OH - ] = 6.13 X 10 -7 M

Ch. 15 Review/Warm-up # 47 If [OH - ] of an aqueous solution is 0.0134 mol/L, what is the pH? pH = -log [H 3 O + ] [H 3 O + ] [OH - ] = 1.00 X 10 -14 M [H 3 O + ] (0.0134 M) = 1.00 X 10 -14 M [H 3 O + ] = 7.46 X 10 -13 M pH = -log (7.46 X 10 -13 M) pH = 12.1

Ch. 15 Review/Warm-up #49 LiOH is a strong base. What is the pH of a 0.082 M LiOH solution? pH = -log [H 3 O + ] LiOH Li + + OH - 0.082 M 0.082 M [H 3 O + ] [OH - ] = 1.00 X 10 -14 M [H 3 O + ] (0.082 M) = 1.00 X 10 -14 M [H 3 O + ] = 1.2 X 10 -13 M pH = -log (1.2 X 10 -13 ) pH = 13 (2 sig figs)

Ch. 15 Review/Warm-up # 63 What is the hydroxide ion concentration in a solution of pH 8.72? [H 3 O + ] = 10 -pH [H 3 O + ] = 10 -8.72 [H 3 O + ] = 1.91 X 10 -9 M [H 3 O + ] [OH - ] = 1.00 X 10 -14 M 1.91 X 10 -9 M [OH - ] = 1.00 X 10 -14 M [OH - ] = 5.24 X 10 -6 M

Ch. 15 Review/Warm-up #78 If 50.00 mL of 1.000 M HI is neutralized by 35.41 mL of KOH, what is the molarity of the KOH? HI + H 2 O I - + H 3 O + 1.000 M 1.000 M (C OH- ) (V OH- ) = (C H3O+ ) (V OH- ) (C OH- ) (0.03541 L) = (1.000 M) (0.05000 L) (C OH- ) = 1.412 M KOH K+ + OH- 1.412 M 1.412 M

Ch. 15 Test “Acids and Bases”  You may use your notes, a periodic table, ion charts, and a calculator.

Download ppt "Thursday, May 15 th : “A” Day Friday, May 16 th : “B” Day Agenda  Collect Crossword Puzzle  Collect Concept Review  Ch. 15 Review – pass back/warm-up."

Similar presentations

Ads by Google