1. Which of the following aqueous solutions would conduct electricity?
AgNO3
C11H22O11 (sucrose)
CH3CH2OH (ethanol)
All of the above
None of the above

2. Which of the following aqueous solutions would conduct electricity?
AgNO3
C11H22O11 (sucrose)
CH3CH2OH (ethanol)
All of the above
None of the above
Answer: a

3. Your Turn Your Turn Consider the following reaction :
Na2SO4(aq) + BaCl2(aq) → 2NaCl(aq) + BaSO4(s)
Write the correct net ionic equation.
2Na+(aq) + SO42–(aq) + Ba2+(aq) + Cl22–(aq) → Na+(aq) + 2Cl–(aq) + BaSO4(s)
B. 2Na+(aq) + SO42–(aq) + Ba2+(aq) + 2Cl–(aq) → 2Na+(aq) + 2Cl–(aq) + BaSO4(s)
C. 2Na+(aq) + SO42–(aq) + Ba2+(aq) + Cl22–(aq) → 2Na+(aq)+ 2Cl–(aq) + Ba2+(s) + SO42–(s)
D. Ba2+(aq) + SO42–(aq) → BaSO4(s)
E. Ba2+(aq) + SO42–(aq) → Ba2+(s) + SO42–(s)

4. Learning Check: Convert Ionic Equation to Net Ionic Equation
Write the correct net ionic equation for each.
Pb2+(aq) + 2NO3–(aq) + 2K+(aq) + 2IO3–(aq) →
Pb(IO3)2(s)+ 2K+(aq) + 2NO3–(aq)
2Na+(aq) + 2Cl–(aq) + Hg22+(aq) + 2NO3–(aq) →
2Na+(aq)+ 2NO3–(aq) + Hg2Cl2(s)
Pb2+(aq) + 2IO3–(aq) → Pb(IO3)2(s)
Note that Hg22+ is a polyatomic ion.
2Cl–(aq) + Hg22+(aq) → Hg2Cl2(s)

5. Predicting Precipitation Reactions
Write the formulas of the compounds being mixed:
K2CO3(aq) + NiCl2(aq) 
2) Write formulas of possible products by mixing components:
KCl NiCO3
3) Consult solubility rules: KCl soluble; NiCO3 insoluble
4) If all products are soluble write: No Reaction N/A
5) If any products are insoluble write their formulas as
products of the reaction using (s) to indicate solid.
K2CO3(aq) + NiCl2(aq) NiCO3(s) + KCl(aq)
6) Balance the equation adjusting only the coefficients
and not the subscripts:
K2CO3(aq) + NiCl2(aq) NiCO3(s) + 2KCl(aq)

6. Predict the Products Pb(NO3)2(aq) + Ca(OH)2(aq) 
BaCl2(aq) + Na2CO3(aq) 
2Na3PO4(aq) + 3Hg2(NO3)2(aq) 
2 NaCl(aq) + Ca(NO3)2(aq) 
Pb(OH)2(s) + Ca(NO3)2(aq)
BaCO3(s) + 2NaCl(aq)
6NaNO3(aq) + (Hg2)3(PO4)2(s)
Note that Hg22+ is a polyatomic ion.
NR (No reaction)
CaCl2(aq) + 2NaNO3(aq)

7. Predicting Reactions and Writing Their Equations
What reaction, if any, occurs between potassium nitrate and ammonium chloride?
Need to know whether net ionic equation exists.
Determine formulas of reactants
KNO3 + NH4Cl  ?
Write molecular equation
KNO3 + NH4Cl  KCl + NH4NO3
Check solubilities
All are soluble

8. Predicting Reactions and Writing Their Equations
Predicted molecular equation
KNO3(aq) + NH4Cl(aq)  KCl(aq) + NH4NO3(aq)
Write ionic equation
K+(aq) + NO3–(aq) + NH4+(aq) + Cl–(aq)  K+(aq) + Cl–(aq) + NH4+(aq) + NO3–(aq)
Same on both sides
All ions cancel out
No gases, solids, water, or weak electrolytes formed – NO REACTION

9. Predict the products and balance the equation.
KCl(aq) + AgNO3(aq) ®
Na2S(aq) + CaCl2(aq) ®
(K+ + Cl−) + (Ag+ + NO3−) → (K+ + NO3−) + (Ag+ + Cl−)
Most halides are soluble - AgCl is the exception.
Reaction is already balanced.
KCl(aq) + AgNO3(aq) → KNO3 + AgCl
KCl(aq) + AgNO3(aq) ® KNO3(aq) + AgCl(s)
(Na+ + S2−) + (Ca2+ + Cl−) → (Na+ + Cl−) + (Ca2+ + S2−)
Na2S(aq) + CaCl2(aq) → NaCl + CaS
most sulfides are insoluble - CaS is the exception (partially soluble).
Na2S(aq) + CaCl2(aq) ® NaCl(aq) + CaS(aq)
No reaction: STOP!

10. Neutralization Reactions
An acid and base can neutralize each other and form a salt and water.
Note aq for salt
acid + base salt(aq) + water
HBr (aq) + KOH (aq) KBr (aq) + H2O
H+(aq) + Br-(aq) + K+(aq) + OH-(aq) K+(aq) + Br-(aq) + H2O
Net ionic equation: H+ (aq) + OH- (aq) H2O

11. Solution Stoichiometry
If you dissolve 25.5 g KBr in enough water to make 1.75 L of solution, what is the molarity of the solution?
Calculate moles of KBr = 25.5 g/( ) = mol
Calculate molarity = 0.214/1.75 L = M
Note the capital M refers to moles/liter

12. Solution Stoichiometry
g of calcium chloride is dissolved in enough water to make a mL aqueous solution. What is the molarity of this aqueous calcium chloride solution ?

13. How many liters of 0.125 M NaOH contain 0.255 mol NaOH?
Given:
Find:
0.125 M NaOH, mol NaOH
liters, L
Conceptual Plan:
Relationships:
0.125 mol NaOH = 1 L solution
mol NaOH
L sol’n
Solution:
A useful formula to remember is liters X Molarity = moles
Check:
because each L has only mol NaOH, it makes sense that mol should require a little more than 2 L

14. How would you prepare 250. 0 mL of a 1
How would you prepare mL of a 1.00 M solution of CuSO45 H2O (MM )?
Given:
Find:
250.0 mL solution; 1.00 M, MM =
mass CuSO4 5 H2O, g
Conceptual Plan:
Relationships:
1.00 L sol’n = 1.00 mol; 1 mL = L; 1 mol = g
mL sol’n
L sol’n
g CuSO4
mol CuSO4
Solution:
Dissolve 62.4 g of CuSO4∙5H2O in enough water to total mL
Check:
the unit is correct, the magnitude seems reasonable as the volume is ¼ of a liter

15. Using Molarity in Calculations
copper(II) sulfate pentahydrate is a beautiful blue compound that is highly water soluble. Describe the preparation of mL of M aqueous copper(II) sulfate solution ?
M x L = moles

16. Using Molarity in Calculations
Weigh out 19.4 g CuSO4∙5H2O and dissolve in enough DI water to form a total of mL of solution. Mix well and serve. The resulting solution is M CuSO4 (aq)

17. Preparing 3.00 L of 0.500 M CaCl2 from a 10.0 M Stock Solution

18. To what volume should you dilute 0. 200 L of 15. 0 M NaOH to make 3
To what volume should you dilute L of 15.0 M NaOH to make 3.00 M NaOH?
Given:
Find:
V1 = 0.200L, M1 = 15.0 M, M2 = 3.00 M
V2, L
Conceptual Plan:
Relationships:
M1V1 = M2V2
V1, M1, M2 V2
Solution:
Check:
because the solution is diluted by a factor of 5, the volume should increase by a factor of 5, and it does

19. Describe the preparation of 400. 0 mL of 0
Describe the preparation of mL of M C12H22O11 (aq) solution from a 1.00 M C12H22O11 (aq) stock solution.
obtain 88.0 mL of 1.00 M C12H22O11 (aq)
add enough water to form a total, final volume of mL (that is, add mL of water to the 88.0 mL of 1.00 M C12H22O11 (aq)

20. Solution Stoichiometry
What volume in L of M KCl solution will completely react with L of a M Pb(NO3)2 solution according to the following balanced chemical equation?
2KCl + Pb(NO3)2  PbCl2 + 2 KNO3
Calculate moles of Pb(NO3)2 using L X M = moles
= L X M = moles
Recognize that stoichiometry has 2 KCl : 1 Pb(NO3)2
You need 2 X moles of KCl
Solve for L = moles/M = mol/0.150 M = L

21. What is the maximum number of grams of PbI2 precipitated upon mixing 25.0 mL of M KI with 15.0 mL of M Pb(NO3)2?
0.864 g
1.73 g
1.21 g
2.07 g
None of the above

22. What is the maximum number of grams of PbI2 precipitated upon mixing 25.0 mL of M KI with 15.0 mL of M Pb(NO3)2?
0.864 g
1.73 g
1.21 g
2.07 g
None of the above
Answer: a

23. Solution Stoichiometry
How many milliliters of M H3PO4 could be completely neutralized by 45.0 mL of M KOH? The balanced equation for the reaction is H3PO4(aq) + 3KOH(aq)  K3PO4(aq) + 3H2O Strategy:
Coefficients of
Balanced equation
mol KOH
mol H3PO4
Vol and M of
KOH soln
mol and M of
H3PO4 soln
KOH solution
H3PO4 soln

24. Learning Check: Solution Stoichiometry
1. Calculate moles of KOH 2. Use coefficients to calculate the moles H3PO4 required 3. Calculate volume of H3PO4 needed
= 4.50 × 10–3 mol KOH
= 1.50 × 10–3 mol H3PO4
= 31.6 mL H3PO4

25. A 25. 0 mL sample of H2SO4 is neutralized with NaOH
A 25.0 mL sample of H2SO4 is neutralized with NaOH. What is the concentration of the H2SO4 if 35.0 mL of M NaOH are required to completely neutralize the acid?
0.210 M
0.105 M
0.150 M
0.420 M
None of the above

26. A 25. 0 mL sample of H2SO4 is neutralized with NaOH
A 25.0 mL sample of H2SO4 is neutralized with NaOH. What is the concentration of the H2SO4 if 35.0 mL of M NaOH are required to completely neutralize the acid?
0.210 M
0.105 M
0.150 M
0.420 M
None of the above
Answer: b

27. Your Turn
Which of the following is a weak acid?
HCl (hydrochloric acid)
HNO3(nitric acid)
HClO4 (perchloric acid)
HC2H3O2 (acetic acid)
H2SO4 (sulfuric acid)

28. Your Turn
Which of the following is a strong acid?
HF (hydrofluoric acid)
HClO3 (chloric acid)
H3PO4 (phosphoric acid)
HNO2 (nitrous acid)
H2SO3 (sulfurous acid)

29. Write the equation for the process that occurs when the following strong electrolytes dissolve in water.
CaCl2
HNO3
(NH4)2CO3
CaCl2(aq)  Ca2+(aq) + 2 Cl−(aq)
HNO3(aq)  H+(aq) + NO3−(aq)
(NH4)2CO3(aq)  2 NH4+(aq) + CO32−(aq)

30. Learning Check
Write the equations that illustrate the dissociation of the following salts:
Na3PO4(aq) →
Al2(SO4)3(aq) →
CaCl2(aq) →
Ca(MnO4)2(aq) →
3Na+(aq) + PO43–(aq)
2Al3+(aq) + 3SO42–(aq)
Ca2+(aq) + 2Cl–(aq)
Ca2+(aq) + 2MnO4–(aq)

31. Your Turn
How many ions form on the dissociation of Na3PO4? 1 2 3 4 8

32. Your Turn
How many ions form on the dissociation of
Al2(SO4)3? 2 3 5 9 14

33. Which of the following compounds will be insoluble in water?
Ca(OH)2
Na3PO4
CaS
Hg2Cl2
LiF

34. Which of the following compounds will be insoluble in water?
Ca(OH)2
Na3PO4
CaS
Hg2Cl2
LiF
Answer: d

35. What is the molarity of an aqueous solution containing 22
What is the molarity of an aqueous solution containing 22.5 g of sucrose (C12H22O11) in 35.5 mL of solution? M
1.85 × 10–3 M
1.85 M
3.52 M
0.104 M

36. What is the molarity of an aqueous solution containing 22
What is the molarity of an aqueous solution containing 22.5 g of sucrose (C12H22O11) in 35.5 mL of solution? M
1.85 × 10–3 M
1.85 M
3.52 M
0.104 M
Answer: c

37. What volume of 12.1 M HCl is needed to create 250. mL of 3.2 M HCl?
Your Turn
What volume of 12.1 M HCl is needed to create 250. mL of 3.2 M HCl?
66 mL
800 mL
3025 mL
945 mL
9680 mL
Vconc = 66 mL

38. Your Turn
A 25 mL of 6.0 M HCl is diluted to 500 mL with water. What is the molarity of the resulting solution?
150 M
3.0 M
0.120 M
120 M
0.30 M
Mdil = 0.30 M

39. Sulfuric acid is found in some types of batteries. What volume of 3
Sulfuric acid is found in some types of batteries. What volume of 3.50 M H2SO4 is required to prepare mL of 1.25 M H2SO4?
17.5 mL
700. mL
89.3 mL
109 mL
None of the above

40. Sulfuric acid is found in some types of batteries. What volume of 3
Sulfuric acid is found in some types of batteries. What volume of 3.50 M H2SO4 is required to prepare mL of 1.25 M H2SO4?
17.5 mL
700. mL
89.3 mL
109 mL
None of the above
Answer: c

41. A 45.0 mL solution is M HCl. What is the concentration after 25.0 mL of water is added to the solution? Assume volumes are additive.
0.321 M
0.179 M
0.900 M
0.278 M

42. A 45.0 mL solution is M HCl. What is the concentration after 25.0 mL of water is added to the solution? Assume volumes are additive.
0.321 M
0.179 M
0.900 M
0.278 M
Answer: a

43. Your Turn
What will be the solid product of the reaction of Ca(NO3)2(aq) + Na2CO3(aq)  ?
CaCO3
NaNO3
Na(NO3)2
Na2(NO3)2
H2O

44. What is the product, if any, when AgNO3 and NaCl react in aqueous solution?
AgCl
NaNO3
No precipitate

45. What is the product, if any, when AgNO3 and NaCl react in aqueous solution?
AgCl
NaNO3
No precipitate
Answer: c

46. Which of the following is not electrically balanced?
Your Turn
Which of the following is not electrically balanced?
A. Ag+(aq) + Cl–(aq)  AgCl(s)
B. NH4+(aq) + H2O  NH3(aq) + H3O+(aq)
C. NO2–(aq) + H2O  HNO2(aq) + OH–(aq)
D. Mg+(aq) + 2OH–(aq)  Mg(OH)2(s)
E. Cu2+(aq) + Sn(s)  Cu(s) + Sn2+(aq)

47. Which of the following is not electrically balanced?
Your Turn
Which of the following is not electrically balanced?
A. PO43-(aq) + H+(aq)  HPO42-(aq)
B. 2Ag+(aq) + Zn(aq)  Zn2+(aq) + 2Ag(s)
C. H2PO4–(aq)  2H+(aq) + PO4(aq)
D. CO32-(aq) + 2H+(aq)  H2CO3(aq)
E. Pb2+(aq) + S2-(aq)  PbS(s)

48. Learning Check: Convert Molecular to Ionic Equations:
Write the correct ionic equation for each:
Pb(NO3)2(aq) + 2NH4IO3(aq) →
Pb(IO3)2(s) + 2NH4NO3(aq)
2NaCl (aq) + Hg2(NO3)2 (aq) → 2NaNO3 (aq) + Hg2Cl2 (s)
Pb2+(aq) + 2NO3–(aq) + 2NH4+(aq) + 2IO3–(aq) → Pb(IO3)2(s) + 2NH4+(aq) + 2NO3–(aq)
Note that Hg22+ is a polyatomic ion.
2Na+(aq) + 2Cl–(aq) + Hg22+(aq) + 2NO3–(aq) → 2Na+(aq) + 2NO3–(aq) + Hg2Cl2(s)

49. Consider the following molecular equation:
Your Turn
Your Turn
Consider the following molecular equation:
(NH4)2SO4(aq) + Ba(CH3CO2)2(aq) →
2NH4CH3CO2(aq) + BaSO4(s)
Write the correct net ionic equation.
Ba2+(aq) + SO42–(aq) → BaSO4(s)
2NH4+(aq) + 2CH3CO2–(aq) → 2NH4CH3CO2(s)
Ba2+(aq) + SO42–(aq) → BaSO4(aq)
2NH4+(aq) + Ba2+(aq) + SO42–(aq) + 2CH3CO2–(aq) → NH4+(aq) + 2CH3CO2–(aq) + BaSO4(s)
2NH4+(aq) + 2CH3CO2–(aq) → 2NH4CH3CO2(aq)

50. Write the ionic and net ionic equation for each of the following:
Examples
Write the ionic and net ionic equation for each of the following:
1. K2SO4(aq) + 2 AgNO3(aq) ® 2 KNO3(aq) + Ag2SO4(s)
2 K+(aq) + SO42−(aq) + 2 Ag+(aq) + 2 NO3−(aq)
® 2 K+(aq) + 2 NO3−(aq) + Ag2SO4(s)
2 Ag+(aq) + SO42−(aq) ® Ag2SO4(s)
2. Na2CO3(aq) + 2 HCl(aq) ® 2 NaCl(aq) + CO2(g) + H2O(l)
2 Na+(aq) + CO32−(aq) + 2 H+(aq) + 2 Cl−(aq)
® 2 Na+(aq) + 2 Cl−(aq) + CO2(g) + H2O(l)
CO32−(aq) + 2 H+(aq) ® CO2(g) + H2O(l)

51. Your Turn
What is the net ionic equation for the reaction: NaOH(aq) + HF(aq)  ? A. Na+(aq) + OH–(aq) + H+(aq) + F–(aq)  H2O + NaF(aq) B. OH–(aq) + H+(aq)  H2O C. Na+(aq) + OH–(aq) + HF(aq)  H2O + NaF(aq) D. OH–(aq) + HF(aq)  H2O + F–(aq) E. No reaction

52. What is the net ionic equation for the reaction:
Your Turn
What is the net ionic equation for the reaction:
NH3(aq) + CH3COOH(aq)  ?
A. NH2–(aq) +H+(aq) + CH3COOH(aq)  NH3(aq) HCH3COOH(aq)
B. NH3(aq) + CH3COO–(aq) + H+(aq)  NH4+(aq) CH3COO–(aq)
C. NH3(aq) + CH3COO–(aq) + H+(aq)  NH4CH3COO(s)
D. NH3(aq) + CH3COOH(aq)  NH4+(aq) + CH3COO– (aq)
E. No reaction

53. Your Turn
Which of the following is not a strong base?
NaOH
CH3NH2
Cs2O
Ba(OH)2
CaO

54. Your Turn
Which of the following is an acid?
NaO2
SO2
C. CH3NH2
Ba(OH)2
CaO

55. Which of the following is a base? N(CH3)3 SO2 C. CH3COOH D. HF E. HNO2
Your Turn
Which of the following is a base?
N(CH3)3
SO2
C. CH3COOH
D. HF
E. HNO2
N(CH3)3(aq) + H2O  HN(CH3)3+(aq) + OH–(aq)
Just like ammonia, NH3
NH3(aq) + H2O  NH4+(aq) + OH–(aq)

56. Your Turn
What is the correct name for HClO4 (aq)?
chloric acid
hydrochloric acid
perchloric acid
hypochlorous acid
chlorous acid

57. Your Turn
What is the correct name for H2SO3(aq)?
sulfuric acid
sulfurous acid
hydrosulfuric acid
hydrosulfurous acid
hydrogen sulfite acid

58. Write the molecular, ionic, and net-ionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide.
1. Write the formulas of the reactants
HNO3(aq) + Ca(OH)2(aq) 
2. Determine the possible products
a) determine the ions present when each reactant dissociates or ionizes
(H+ + NO3−) + (Ca2+ + OH−) 
b) exchange the ions, H+ combines with OH− to make H2O(l)
(H+ + NO3−) + (Ca2+ + OH−)  (Ca2+ + NO3−) + H2O(l)
c write the formula of the salt
(H+ + NO3−) + (Ca2+ + OH−)  Ca(NO3)2 + H2O(l)

59. Write the molecular, ionic, and net-ionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide.
3. Determine the solubility of the salt Ca(NO3)2 is soluble 4. Write an (s) after the insoluble products and an (aq) after the soluble products HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + H2O(l) 5. Balance the equation 2 HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + 2 H2O(l)

60. Write the molecular, ionic, and net-ionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide.
2 HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + 2 H2O(l)
6. Dissociate all aqueous strong electrolytes to get complete ionic equation
not H2O
2 H+(aq) + 2 NO3−(aq) + Ca2+(aq) + 2 OH−(aq) 
Ca2+(aq) + 2 NO3−(aq) + H2O(l)
7. Eliminate spectator ions to get net-ionic equation
2 H+(aq) + 2 OH−(aq)  2 H2O(l)
H+(aq) + OH−(aq)  H2O(l)

61. Predict the products and balance the equation.
HCl(aq) + Ba(OH)2(aq) ®
H2SO4(aq) + Sr(OH)2(aq) ®
(H+ + Cl−) + (Ba2+ + OH−) → (H+ + OH−) + (Ba2+ + Cl−)
most halides are soluble – BaCl2 is soluble
HCl(aq) + Ba(OH)2(aq) → H2O(l) + BaCl2
2 HCl(aq) + Ba(OH)2(aq) → 2 H2O(l) + BaCl2(aq)
(H+ + SO42−) + (Sr2+ + OH−) → (H+ + OH−) + (Sr2+ + SO42−)
H2SO4(aq) + Sr(OH)2(aq) → H2O(l) + SrSO4
some sulfates are soluble, a few (SrSO4) are not
H2SO4(aq) + Sr(OH)2(aq) → 2 H2O(l) + SrSO4
H2SO4(aq) + Sr(OH)2(aq) → 2 H2O(l) + SrSO4(s)

62. One More 1. H2SO4(aq) + LiOH(aq) ®
(H+ + SO42−) + (Li+ + OH−) → (H+ + OH−) + (Li+ + SO42−)
H2SO4(aq) + LiOH(aq) → H2O(l) + Li2SO4
H2SO4(aq) + 2 LiOH(aq) ® 2 H2O(l) + Li2SO4(aq)

63. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Given: mL HCl
12.54 mL of M NaOH,
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Find: M HCl
Plan: ml NaOH → moles NaOH → moles HCl → M HCl
Conversion factors and relationships:
1 mole: HCl = 1 mole NaOH (from reaction)

64. Write a conceptual plan
The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Information
Given: mL HCl
12.54 mL of M NaOH
Find: M HCl
Rel: 1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
Write a conceptual plan mL
NaOH L
NaOH
mol
NaOH
mol
HCl mL
HCl L
HCl

65. Apply the conceptual plan
The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Information
Given: mL HCl
12.54 mL of M NaOH
Find: M HCl
Rel: 1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
CP: mL NaOH → L NaOH →
mol NaOH → mol HCl;
mL HCl → L HCl & mol  M
Apply the conceptual plan
= 1.25 x 103 mol HCl

66. Apply the conceptual plan
The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Information
Given: mL HCl
12.54 mL NaOH
Find: M HCl
Rel: 1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
CP: mL NaOH → L NaOH →
mol NaOH → mol HCl;
mL HCl → L HCl & mol  M
Apply the conceptual plan

67. Check the solution HCl solution = 0.125 M
The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Information
Given: mL HCl
12.54 mL NaOH
Find: M HCl
Rel: 1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
CP: mL NaOH → L NaOH →
mol NaOH → mol HCl;
mL HCl → L HCl & mol  M
Check the solution
HCl solution = M
The units of the answer, M, are correct.
The magnitude of the answer makes sense because
the neutralization takes less HCl solution than
NaOH solution, so the HCl should be more concentrated.

68. What mass, in grams, of sodium bicarbonate, NaHCO3, is required to neutralize L of M H2SO4?
2.94 × 104 g
1.47 × 104 g
5.88 × 104 g
3.50 × 103 g
1.75 × 103 g

69. What mass, in grams, of sodium bicarbonate, NaHCO3, is required to neutralize L of M H2SO4?
2.94 × 104 g
1.47 × 104 g
5.88 × 104 g
3.50 × 103 g
1.75 × 103 g
Answer: c

70. Your Turn
A mL sample of HNO3 is titrated with mL of 1.30 M Ca(OH)2. What is the concentration of HNO3 in the initial sample?
2HNO3(aq) + Ca(OH)2(aq)  2AgBr(s) + Ca(NO3)2(aq)
0.433 M
1.95 M
0.867 M
3.90 M
7.80 M
= 7.80 M HNO3