Chapter 8: Solutions, Molarity, and Reactions

Solute (solid, liquid, gas)

A __________ is the substance dissolved in the solvent

Solvent (aqueous solution)

The ___________ dissolves the solute

Diluted: low concentration

Concentrated: high concentration

A ____________ solution has low concentration while a _____________ solution has high concentration

Molarity (M) = mol solute/ L solution

What equation is used to solve for molarity (M)

0.250 mol/L NaCl

0.250M NaCl → ? mol/L NaCl

2.125mg x (10^-3 g/ 1 mg) x (1 mol/58.443g) = 3.636 x 10^-5 mol

(3.636 x 10^-5 mol)/ 1.000 L = 3.636 x 10^-5 M

3.636 x 10^-5 M x (1 mM/ 10^-3M) = 3.636 x 10^-2 mM

Solve for Molarity:

2.125mg NaCl is dissolved to make a 1.000L solution

0.1000 mol NaCl x (1 L solution/ 3.636 x 10^-5 mol NaCl)

= 2750. L ← needed

Need 0.1000 mol NaCl. What volume of the 3.636 x 10^-5 M solution should be used?

250.0mL x (10^-3 L/ 1mL) x (0.0800 mol Na2CrO4/ 1 L solution) x (161.97g Na2CrO4/ 1 mol Na2CrO4

= 3.24g Na2CrO4

Take 3.24g Na2CrO4 and add enough water to make a 250.0mL solution

Preparing solutions:

How would you prepare 250. mL of a 0.0800 M Na2CrO4 solution

1. Find liters

500.0mL x (10^-3 L/ 1mL) = 0.5 L

2. Find moles

1.50g KCl x (1 mol KCl/ 74.55g KCl) = 0.0201 mol

3. Find molarity

0.0201 mol KCl/ 0.5 L solution = 0.0402 M

Solve for Molarity:

1.50g KCl (74.55 g/mol) and 500.0 mL solution is given

0.222 mol NaNO3 x (1 L solution/ 0.500 mol NaNO3) x (1mL/ 10-3 L solution) = 444mL

0.222 mol NaNO3 and 0.500 M solution is needed, find the volume in mL

True

True or False:

Molarity decreases as a highly concentrated solution is diluted with water

(Minitial) x (Vinitial) = (Mfinal) x (Vfinal)

M - Molarity

V - Volume

What does this equation mean:

M1V1 = M2V2

True

True or False:

The volume (V) amount in M1V1 = M2V2 can use any unit as long as it is the same on each side of the equation

M1V1 = M2V2

(12.0M) x V1 = (0.500M) x (100.mL)

V1 = 4.17 mL

4.17mL of 12.0M solution is used to dilute a total of 100.mL

12.0M HCl (M1) is bought. I need 100mL (V2) of a 0.500M HCl (M2) solution. How much of the 12.0 M solution is needed?

(0.50M) x (100.mL) = (0.10M) x V2

V2 = 500.mL

What is the total volume (V2) of H2O needed to dilute 100.mL (V1) of 0.50M (M1) to reach 0.10M (M2)

Electrolytes

_____________ are substances that dissociate into ions in a solution

Nonelectrolytes

_____________ are substances that do not dissociate into ions in a solution

Weak electrolytes partially dissociate into ions

strong electrolytes completely dissociate into ions

What is the difference between weak electrolytes and strong electrolytes/strong acids?

Nonelectrolyte, does not dissociate any further

Is H2O (l) an electrolyte (weak or strong) or a nonelectrolyte?

Strong electrolyte, dissociates into NaCl (s) → Na+ (aq) + Cl- (aq)

Is NaCl (aq) an electrolyte (weak or strong) or a nonelectrolyte?

K3PO4(s) → 3K+(aq) + PO4^3-(aq)

What is the dissociation equation of K3PO4(aq)

True

True or False:

When any ionic compound dissolves into water, the compound dissociates into ions

H2O

CH3

CH2

OH

Most molecular compounds do not form ions in water such as

_________

_________

_________

_________

Weak electrolyte, partially dissociates into ions

Is acetic acid (CH2COOH) a weak electrolyte or a strong electrolyte?

Acids → proton (H+) donors

Bases → proton acceptors

According to the Bronsted - Lowry definition:

Acids are _____________ and bases are ____________

Monoprotic acids (1)

Diprotic acid (2)

Acids that donate one proton (H+) are ______________

Acids that donate two protons (H+) are ______________

1. perchloric acid (HClO4)

2. chloric acid (HClO3)

3. hydrochloric acid (HCl)

4. hydrobromic acid (HBr)

5. hydroiodic acid (HI)

6. nitric acid (HNO3)

7. sulfuric acid (H2SO4)

What are the 7 strong acids you need to recognize

Monoprotic acids (1)

1. perchloric acid (HClO4)

2. chloric acid (HClO3)

3. hydrochloric acid (HCl)

4. hydrobromic acid (HBr)

5. hydroiodic acid (HI)

6. nitric acid (HNO3)

Diprotic acid (2)

7. sulfuric acid (H2SO4)

Out of the 7 strong acids you need to recognize, which are monoprotic acids and which are diprotic acids

HCl (aq) → acid

H2O (l) → base

One arrow (→) indicates 100% of ions are in the product (completely ionized)

Identify the base and acid. What does the single arrow indicate?

HCl (aq) + H2O (l) → H3O+(aq) + Cl-(aq)

1 and 2

Strong bases are found in Groups ____ and ___ on the periodic table

False

Group 1 completely dissolves

Group 2 does not dissolve

True or False:

Strong bases found in Group 2 completely dissolve into water while strong bases found in Group 1 do not dissolve

Group 1 - (aq) because they are soluble

Group 2 - (s) because they are not soluble

What state of matter is assigned to the strong bases found in Group 1 and Group 2 and why

They are all Group 1 Bases

What do these formulas have in common

KOH

NaOH

LiOH

CsOH

They are all Group 2 Bases

What do these formulas have in common

Ca(OH)2

Mg(OH)2

Ba(OH)2

Sr(OH)2

True

True or False:

Forming hydroxide makes the solution basic

NaOH(s) → Na+(aq) + OH–(aq)

NH3(aq) → acid

H2O(l) → base

⇆ indicates partially dissociated ions in the product (partially ionized)

Identify the base and acid. What do the double arrows indicate?

NH3(aq) + H2O(l) ⇆ NH4+(aq) + OH-(aq)

Neutralizing reactions

_______________ reactions occur when an acid and base react to form H2O and an ionic compound

1. Molecular equation - balance

HCl (aq) + NaOH (aq) → H2O(l) + NaCl(aq)

2. Ionic equation - Ionize

H+(aq) + Cl–(aq) + Na+(aq) + OH–(aq) → H2O(l) + Na+(aq) + Cl–(aq)

3. Net ionic equation - Cancel out spectator ions

H+(aq) + Cl–(aq) + Na+(aq) + OH–(aq) → H2O(l) + Na+(aq) + Cl–(aq)

H+(aq) + OH–(aq) → H2O(l)

Neutralizing reactions

HCl (aq) + NaOH (aq) → ?

1. Molecular equation - balance

HNO3(aq) + Mg(OH)2(s) → H2O(l) + Mg(NO3)2(aq)

2 HNO3(aq) + Mg(OH)2(s) → 2 H2O(l) + Mg(NO3)2(aq)

2. Ionic equation - Ionize

2H+(aq) + 2NO3–(aq) + Mg(OH)2(s) → 2 H2O(l) + Mg2+(aq) + 2 NO3–(aq)

3. Net ionic equation - Cancel out spectator ions

2H+(aq) + 2NO3–(aq) + Mg(OH)2(s) → 2 H2O(l) + Mg2+(aq) + 2NO3–(aq)

2H+(aq) + Mg(OH)2(s) → 2 H2O(l) + Mg2+(aq)

Neutralizing reactions

HNO3(aq) + Mg(OH)2 → ?

1. Molecular equation - balance

CH3COOH (aq) + NaOH (aq) → H2O (l) + NaCH3COO (aq)

2. Ionic equation - Ionize

CH3COOH (aq) + Na+(aq) + OH–(aq) → H2O (l) + Na+(aq) + CH3COO–(aq)

3. Net ionic equation - Cancel out spectator ions

CH3COOH (aq) + Na+(aq) + OH–(aq) → H2O (l) + Na+(aq) + CH3COO–(aq)

CH3COOH (aq) + OH–(aq) → H2O (l) + CH3COO–(aq)

Neutralizing reactions

CH3COOH (aq) + NaOH (aq) → ?

True

True or False:

CH3COOH (aq) → weak acid/electrolyte → will not ionize for net ionic equation

Continuum

Range of solubility:

Solubility is on a __________

Solubility rules

___________ determine the state of matter for each variable in a precipitation reaction

1. Molecular equation - balance and determine state of matter

NaI + AgNO3 → NaNO3 + AgI

NaI (aq) + AgNO3 (aq) → NaNO3 (aq) + AgI (s)

2. Ionic equation - Ionize

Na+(aq) + I–(aq) + Ag+(aq) + NO3–(aq) → Na+(aq) + NO3–(aq) + AgI (s)

3. Net ionic equation - Cancel out spectator ions

Na+(aq) + I–(aq) + Ag+(aq) + NO3–(aq) → Na+(aq) + NO3–(aq) + AgI (s)

I–(aq) + Ag+(aq) → AgI (s)

Precipitation reactions:

NaI + AgNO3 → ?

1. Molecular equation - balance and determine state of matter

Ca(NO3)2 + K2SO4 → CaSO4 + KNO3

Ca(NO3)2 (aq) + K2SO4 (aq) → CaSO4 (s) + KNO3 (aq)

Ca(NO3)2 (aq) + K2SO4 (aq) → CaSO4 (s) + 2KNO3 (aq)

2. Ionic equation - Ionize

Ca2+(aq) + 2NO3– (aq) + 2K+(aq) + SO42–(aq) → CaSO4 (s) + 2K+(aq) + 2NO3– (aq)

3. Net ionic equation - Cancel out spectator ions

Ca2+(aq) + 2NO3– (aq) + 2K+(aq) + SO42–(aq) → CaSO4 (s) + 2K+(aq) + 2NO3– (aq)

Ca2+(aq) + SO42–(aq) → CaSO4 (s)

Precipitation reactions:

Ca(NO3)2 + K2SO4 → ?

KI (aq) + NaCl (aq) → KCl (aq) + NaI (aq)

No reaction → no net ionic equation → all soluble (aq)

KI (aq) + NaCl (aq) → ?

NH4NO3 (aq) + KCl (aq) → NH4Cl (aq) + KNO3(aq)

No reaction

NH4NO3 + KCl → ?

1. Molecular equation - balance and determine state of matter

Pb(ClO4)2 + CaCl2 → PbCl2 + Ca(ClO4)2

Pb(ClO4)2 (aq) + CaCl2 (aq) → PbCl2 (s) + Ca(ClO4)2 (aq)

2. Ionic equation - Ionize

Pb2+(aq) 2ClO4–(aq) + Ca2+(aq) + 2Cl–(aq) → PbCl2 (s) + Ca2+(aq) + 2ClO4–(aq)

3. Net ionic equation - Cancel out spectator ions

Pb2+(aq) 2ClO4–(aq) + Ca2+(aq) + 2Cl–(aq) → PbCl2 (s) + Ca2+(aq) + 2ClO4–(aq)

Pb2+(aq) + 2Cl–(aq) → PbCl2 (s)

Precipitation reactions:

Pb(ClO4)2 + CaCl2 → ?

X + Y = XY

No, reduce coefficients to one

Is this reaction finalized?

2X + 2Y = 2XY

1. Find LR

0.0650 L AgNO3 x (0.250 mol AgNO3/ 1 L AgNO3) x (1 mol CaBr2/ 2 mol AgNO3) x (1 L CaBr2/ 0.210 mol CaBr2) = 0.0387 L

= 38.7mL ← needed

35.0mL is given, not enough CaBr2

CaBr2 → LR

2. Find mass of product (AgBr) using given LR

0.0350 L CaBr2 x (0.210 mol CaBr2/ 1 L CaBr2) x (2 mol AgBr/ 1 mol CaBr2) x (187.77g AgBr/ 1 mol AgBr)

= 2.76g AgBr

What mass of AgBr should form from the reaction of 65.0mL of 0.250M AgNO3 and 35.0mL of 0.210M CaBr2

2AgNO3(aq) + CaBr2(aq) → 2AgBr(s) + Ca(NO3)2(aq)

(0.210 mol CaBr2/ 1 L solution) x (1 mol Ca2+/1 mol CaBr2) = (0.210 mol Ca2+/1 L solution)

[Ca2+] = 0.210M

(0.210 mol CaBr2/ 1 L solution) x (2 mol Br–/1 mol CaBr2) = (0.420 mol Br–/1 L solution)

[Br–] = 0.420M

What are the molar concentrations of Ca2+ and Br- in 0.210M CaBr2

CaBr(s) → Ca2+(aq) + 2Br-(aq)

0.420M x (1mM/ 10-3M) = 420. mM

(0.420 mol Br–/ 1 L solution) x (79.904 Br–/ 1 mol Br–) = 33.6 g/L

What is [Br-] in mM and g/L

[Br-] = 0.420M