Acid/Base pH and pOH Solubility Molarity Neutral- ization 100 100 100 100 100 200 200 200 200 200 300 300 300 300 300 400 400 400 400 400 500 500 500 500 500

This is a H+ acceptor. A. Acid B. Base C. Neutral ?1 $100 This is a H+ acceptor. A. Acid B. Base C. Neutral B

This turns red litmus paper blue. A. Acid B. Base C. Neutral ?1 $200 This turns red litmus paper blue. A. Acid B. Base C. Neutral B

Ca(OH)2 A. Acid B. Base C. Neutral ?1 $300 Ca(OH)2 A. Acid B. Base C. Neutral B

?1 $400 pH of 1.5 (Use strong or weak in your response) A. Strong acid B. Weak acid C. Strong base D. Weak base E. Neutral A

A H+ donator. A. Acid B. Base C. Neutral ?1 $500 A H+ donator. A. Acid B. Base C. Neutral A

?2 $100 pH of 0.001 HCl A. 1 B. 3 C. 11 D. 14 B

?2 $200 pOH of 0.00001 M NaOH A. 1 B. 5 C. 9 D. 14 B

?2 $300 pH of 0.01 M Ca(OH)2 A. 1 B. 2 C. 12 D. 14 C

?2 $400 pOH of 0.00001 M HCl A. 1 B. 5 C. 9 D. 14 C

H+ concentration of water A. 1 x 10-7 B. 1 x 10-14 C. 7 D. 14 ?2 $500 H+ concentration of water A. 1 x 10-7 B. 1 x 10-14 C. 7 D. 14 A

?3 $100 Cooling does this to solubility. A. Increases B. Decreases C. No Change B

?3 $200 To dilute a solution you should add more of this. A. Solute B. Solvent C. Solution B

This is not changed in a dilution. ?3 $300 This is not changed in a dilution. Original Diluted A. mL of water 50mL 100mL B. Mass of solution 100g 150g C. Moles of solute 2.0 C

?3 $400 A solution where no more solute molecules can dissolve. (it is just right!) A. Unsaturated B. Saturated C. Supersaturated B

?3 $500 Using this solubility curve: this is the number of grams of NH4Cl that can dissolve in 100 g of water at 70°. A. 60 g B. 74 g C. 90 g D. cannot tell from this graph A

?4 $100 This is the concentration of a solution with 1.0 moles in 500mL of solution. A. 0.002 M B. 0.5 M C. 1 M D. 2 M E. 500 M D

?4 $200 This is the effect on Molarity when more solvent is added to the solution. A. Increases B. Decreases C. No change B – larger denominator when the solvent (L) increases

?4 $300 What is the concentration of 2.0 L of solution that has 50.0g of NH3? A. 1.5 M B. 25 M C. 425 M D. 1500 M A

?4 $400 This is the number of grams of NH3 needed to make 3.00L of a 0.500 M solution. A. 0.0255 g B. 2.83 g C. 25.5 g D. 102 g C

?4 $500 This is the volume of 18M Hydrochloric acid needed to make 0.100L of 2M solution. A. 0.01 L B. 0.9 L C. 180 L D. 360 L A

?5 $100 This is the process we use in the lab to reach neutralization. A. Molarity B. Dilution C. Titration D. Agitation C

?5 $200 This is the number of moles of HCl needed to neutralize 6 moles of NaOH. A. 6 moles B. 12 moles C. 18 moles A

?5 $300 This is the balanced equation when HCl and NaOH react. A. HCl + NaOH  H2O + NaCl B. HCl + H2O  Cl- + H3O+ C. HCl + H2O  H2O + NaCl A

?5 $400 What is the concentration of sulfuric acid (H2SO4) when 25mL are used to neutralize 30mL of 0.20M sodium hydroxide (NaOH). A. 0.12 M B. 0.24 M C. 0.48 M D. 0.83 M A

?5 $500 You have two un-labled HCl solutions: 0.15M and 1.5M. You use 0.020L of acid to neutralize 0.5M NaOH. What statement is true. A. If you used 0.06L of base the acid is 1.5M. B. If you use 0.006L of base the acid is 1.5M. C. If you used 0.06L of base the acid is 0.15 M. D. An equal amount of base will neutralize 0.15 M and 1.5 M solutions A