Acid and Base Review Game Chemistry

Name the Acid  HBr

Name the Acid  HBr  Hydrobromic Acid

Write the Formula  Lithium hydroxide

Write the Formula  Lithium hydroxide  LiOH

Name the Base  Mg(OH) 2

Name the Base  Mg(OH) 2  Magnesium Hydroxide

Arrhenius  What is the “technical” definition of an acid in regards to ?

Arrhenius  A substance which produces H + when put in water

Write the Formula  Hydroiodidic Acid

Write the Formula  Hydroiodidic Acid  HI

Arrhenius  What is the “technical” definition of a base

Arrhenius  What is the “technical” definition of a base?  A substance which produces OH - when put in water

Name the Base  KOH

Name the Base  KOH  Potassium Hydroxide

 If a soap has a hydrogen ion concentration of 2.0 x M, what is the pH of the solution? Is it an acid or a base?

 [H + ] = 2 x M pH = -log(2 x ) = 5.7  Acid because pH < 7

 What is the hydroxide ion concentration of a solution with a pH of 2.3? Is the substance an acid or base?

 pOH = 14 – 2.3 = 11.7  [OH - ] = = 2 x M  Acid because pH < 7

Name the Acid  H 2 SO 4

Name the Acid  H 2 SO 4  Sulfuric Acid

 What is the hydrogen ion concentration of a sample of phosphoric acid that has a pH of 4.9?

 [H + ] = = 1.3 x M

Neutralization Reaction  Write the balanced neutralization reaction for nitrous acid reacting with potassium hydroxide?

Neutralization Reaction  Write the balanced neutralization reaction for nitrous acid reacting with potassium hydroxide?  HNO 2 + KOH  KNO 2 + H 2 O

Name the Base  Ca(OH) 2

Name the Base  Ca(OH) 2  Calcium Hydroxide

Neutralization Reaction  Write the complete balanced equation for the neutralization of phosphoric acid with calcium hydroxide

Neutralization Reaction  Write the complete balanced equation for the neutralization of phosphoric acid with calcium hydroxide  2H 3 PO 4 + 3Ca(OH) 2  Ca 3 (PO 4 ) 2 + 6H 2 O

 How many grams of copper (II) sulfate pentahydrate will be needed to make 75 mL of a M solution? (The molar mass of copper (II) sulfate pentahydrate is g/mole)

 C M = n/V .25 = n/0.075  n = mol x = 4.7 g

Name the Acid  HNO 2

Name the Acid  HNO 2  Nitrous Acid

 How many kilograms of sucrose, C 12 H 22 O 11,will be needed to make 3.50 L of a 1.15 M solution?

 C M = n/v  1.15 = n/3.5  n = mol x 342 = g / 1000 = 1.38 kg

 What is the pH of a solution made with 0.15 grams of sodium hydroxide in 4500 mL of water?

 0.15 g / 40 = moles  M = /4.5 = 8.33X10-4 [OH-]  pOH = -log (8.33x10-4)  pOH = 3.08  pH = = 10.92

Write the Formula  Phosphoric Acid

Write the Formula  Phosphoric Acid  H 3 PO 4

 What is the molarity of a solution that contains moles of NaHCO 3 in a volume of 115 mL

 C M = n/v  C M = 0.075/.115 = 0.65 M

Write the Formula  Nitric Acid

Write the Formula  Nitric Acid  HNO 3

 A solution has 3.00 moles of solute in 2.00 L of solution, what is its molar concentration? How many moles would there be in 350 mL of solution?

 C M = n/V  C M = 3/2 = 1.5M  1.5 = n/0.35  n = 0.53 mol

Write the Formula  Sodium carbonate

Write the Formula  Sodium carbonate  Na 2 CO 3

 Describe how you would prepare 1.00L of a 0.85 M solution of formic acid HCO 2 H?

 C M = n/v  0.85 = n/1  n = 0.85 mol x 46 = 39.1 g  Take 39.1 grams of formic acid and dissolve in a little bit of water. Put into a 1.00 L volumetric flask and fill to the line with water

 What is the molarity of a sulfuric acid solution which contains 5.4 grams of sulfuric acid in 250 mL of water?

 5.4g/98 = mol  C M = 0.055/0.25 = 0.22 M

 What is the molarity of a potassium hydroxide solution which contains 0.94 moles of potassium hydroxide in 450 mL of water?

 C M = 0.94/0.45 = 2.1 M

Write the Formula  Barium hydroxide

Write the Formula  Barium hydroxide  Ba(OH) 2

 In the titration of 35 mL of liquid drain cleaner containing NaOH, 50 mL of 0.4M HCl must be added to reach the equivalence point. What is the molarity of the base in the cleaner?

 M a V a = M b V b  (0.4)(50) = M b (35) 0.57 M 0.57 M

 Calculate how many milliliters of 0.50M NaOH must be added to titrate 46 mL of 0.40 M HClO 4

 M a V a = M b V b  0.4(46) = 0.5V b  V b = 36.8 mL

Name the Base  NaOH

Name the Base  NaOH  Sodium Hydroxide

 A 15.5 mL sample of M KOH was titrated with an acetic acid solution, It took 21.2 mL of the acid to reach the equivalence point. What is the molarity of the acetic acid?

 (0.215)(15.5) = M a (21.2)  M

 What is the molarity of a solution of acetic acid with a pH of 2.65?

 = M

 A 20 mL sample of an HCl solution was titrated with 27.4 mL of a standard solution of NaOH. The concentration of the standard is M. What is the molarity of the HCl?

 (0.0154)(27.4) = M a (20)  = M

 I want to dilute 20 mL of a 6M solution of acetic acid to a 3.8M solution of acetic acid. How much water should I add to the 6M acetic acid to achieve this?

 6(20) = 3.8V  V = 31.6 mL  ADD 11.6 mL of water

 I mix 20 mL of 4.5M NaCl with 40 mL of water. What is the new concentration of the NaCl?

 4.5(20) = M(60) Note: final volume is mL water added  M = 1.5 M

 A 450 mL solution of 1.5 M HCl is sat out over night. 150 mL of the water evaporated. What is the new concentration of the HCl?

 1.5 (450) = M (300) NOTE: final vol is 450 – 150 evaporated = 300  2.25 M

 What does the term “strong acid” mean?

 Strong acid means that the acid dissociates completely in water

 I used 50 grams of potassium chromate to make a 1.4M solution. What is the volume of the solution?

 50 g K 2 CrO 4 / 194 = 0.26 mol  1.4 =.26/v  V = L

Write the Formula  Acetic Acid

Write the Formula  Acetic Acid  HC 2 H 3 O 2

 What is the pH of a solution made by putting 6.0 grams of phosphoric acid in 250 L of water?

 6 g H 3 PO 4 / 98 = mol  [H 3 PO 4 ] = / 250 = 2.45 x  [H + ] = 3(2.45 x ) = 7.35 x  pH = -log(7.35 x ) = 3.13

Name the Acid  HCl

Name the Acid  HCl  Hydrochloric Acid

 What is the pH of a solution made by putting 3.25 grams of strontium hydroxide in 5000 L of water?

 3.25 g Sr(OH) 2 / = mol  [Sr(OH) 2 ] = / 5000 = 5.34 x M  [OH - ] = 2(5.34 x ) = 1.07 x M  pOH = -log(1.07 x ) = 4.97  pH = 14 – 4.97 = 9.03