AP Chem: Solution Calc review: M= mol
or Liters = mol !
Page 1
!
Liter
M
or mol = (Liters) (M) Dilutions: V1M1 = V2 M2 (units just need to match, ml, L, etc)
1. You need to make 500 mL of a 0.250 M solution of CuSO4. How many grams of CuSO4•5 H2O
do you need? (Include the 5 H2O in the formula weight.) How do you make this solution?
! !
(31.2 g, add water to line)
2. You need to make 100.0 mL of a 1.50 M solution of HCl from some 12.0 M HCl stock solution.
How do you dilute this, assuming you have a 100 mL volumetric flask? !
! ! !
(12.5 mL 12 M HCl, add water to line )
3. You need 0.0200 mol of AgNO3 to react with some copper. How many mL of 0.250 M AgNO3
do you need? !
(80.0 mL)
Many solutions are so dilute that a more convenient size unit is needed:
millimolar is mM = mmol/L !
micromolar is µM = µmol/L
***The Formula Weight of a compound in milligrams / millimole is the same as g/mole
4a. Some Bottled water contains 50.0 milligrams of Magnesium/L of solution. Convert this to
milliMolar (millimol/ L). ( so... convert 50.0 mg of Mg+2 to mmol of Mg+2)!
(2.06 mM)
!
!
b. How many mL of this Magnesium solution would contain 0.100 mmol of Magnesium? !
!
(48.5 mL)
5. You need to make 250.0 mL of a 40.0 mM (millimol/Liter) solution of Mg(NO3)2. How many
grams of Mg(NO3)2 would you need?!
( 1.48 g)
Solution Stoichiometry: Titrations, etc. Example Problems
Titration Shortcut (Only for 1:1 ratio) V1 M1 = V2 M2 or V1 mM1 = V2 mM2 !
(units match)
1. Calcium ions or Magnesium ions react with a titrant called EDTA in a 1: 1 ratio. 10.3 mL of
9.75 mM EDTA solution is used to titrate 15.0 mL of some bottled water. Calculate the mM
(mmol/L) of Calcium ions in the bottled water. !
(6.70 mM)
!
b. How many mg of Calcium would be in 240 mL (a glass) of this bottled water?!
(64.3 mg)
2. And you can mix solutions and grams: How many mL of 15.0 mM EDTA would !
titrate a solution containing 17.2 mg of Ca+2 ? (1:1 ratio)!
Page 2
(28.7 mL)
Shortcut(acid-base only):VaMa(#H)=VbMb(#OH) or mol acid (#H)=mol base (#OH)
3. How many mL of 0.400 M NaOH will neutralize 50.0 mL of 2.00 M H2SO4? !
(500 mL)
Often the Titrant is not very pure. In this case you need to “Standardize” it by first using
another, very pure, solution (the “primary standard”). Once you know the exact Molarty of
your titrant, you can use it to determine the Molarity of an unknown sample.
4a. 10.0 mL of 0.100 M Potassium Hydrogen Pthalate (a “primary standard” with only 1 H) is
neutralized by 9.25 mL of NaOH. Calculate the Molarity of the NaOH.!
(0.108 M )
!
b. Then 13.4 mL of some H2SO4 is titrated with 8.50 mL of the same NaOH. Calculate the
Molarity of the H2SO4. !
(0.0343 M)
What if it’s not a 1: 1 ratio or an Acid Base reaction? You need to use a Mole Ratio, Stoichiometry
and a balanced equation.
5. Use this balanced Redox reaction: 6 KI + KBrO3 + 3 H2O ––> KBr + 3 I2 + 6 KOH
a. How many mL of 0.100 M KI will react with 50.0 mL of 0.250 M KBrO3 ? !
(750 mL)
b. 30.0 mL of 0.400 M KI react with 25.0 mL of KBrO3. Calculate the Molarity of the KBrO3.
! !
(0.0800 M)
c. 15.0 mL of 0.250 M KI react with excess KBRO3. How many grams of I2 would be made?
! !
(0.476 g I2)
6. How many grams of KOH can be neutralized by 100 mL of 0.500 M H3PO4? !
(8.40 g)
AP Chem Solution Stoichiometry: titrations, etc.Practice Problems
page 3
1. a. 14.2 mL of 12.3 mM EDTA are used to titrate 10.0 mL of a Mg(NO3)2 solution. Calculate the
Molarity of the Mg(NO3)2. Mg+2 + EDTA ––––> MgEDTA complex (1:1 ratio)! (17.5 mM)
!
!
b. How many mL of 10.5 mM EDTA would titrate a solution containing 8.75 mg of Mg+2?
! !
(34.3 mL)
2. 4.44 grams of Ca(OH)2 are neutralized by 40.0 mL of some HCl solution. What is the Molarity
of the HCl solution? !
(3.00 M)
3a. Write the Balanced equation for the Double Replacement reaction of Silver nitrate and Sodium
carbonate. Then write the net ionic equation. What is the Insoluble solid in this reaction?
b. How many mL of 0.150 M Na2CO3 solution would react with exactly 25.0 mL of 0.210 M
AgNO3 solution?!
(17.5 mL)
c. 50.00 mL of an unknown solution containing silver ions reacts with excess Na2CO3 . 5.68 g of
the Insoluble solid is produced. Calculate the Molarity of the Silver ions (Ag+) in the unknown
solution.!
(0.824 M)
4. Complete and Balance the Single replacement reaction of Copper(II) chloride solution and
Chromium metal. (Make Chromium (III) ). Then write the Net Ionic equation.
!
!
b. How many grams of Chromium metal would react with 60.0 mL of 0.500 M CuCl2 solution?
!
(1.04 g)
page 4
5. a. Balance this Redox reaction: (Answer below)!
FeCl2 +
KMnO4
+ HCl –––>
FeCl3 +
MnCl2 +
H2O +
KCl
!
b. 16.5 mL of an Fe+2 solution is titrated with 13.2 mL of a 0.0250 M KMnO4 solution.
Calculate the Molarity of the Fe+2 solution.
!
c. A vitamin tablet containing Fe+2 is dissolved in HCl. 11.1 mL of 0.00579 M KMnO4 solution
is required to titrate this amount of iron. How many mg of Fe were in the vitamin tablet?
! !
(a. 5, 1, 8, 5, 1, 4, 1; b. 0.100 M ; c. 17.9 mg Fe)
6. 10.0 mL of an HCl solution reacts with exactly 10.25 mL of 0.0945 M NaOH (to standardize the
HCl). Then an antacid tablet containing Al(OH)3 neutralizes 85.8 mL of the HCl. How many
mg of Al(OH)3 were in the antacid?!
(0.0969 M HCl, 216 mg Al(OH)3)
“Back Titration” Example : An application of the limiting reactant principle.
7. 75.0 mL of 0.200 M HCl reacts with an antacid containing Mg(OH)2 and is allowed to sit
overnight. 10.0 mL of the resulting mixture required 4.50 mL of 0.0950 M NaOH to neutralize.
How many grams of Mg(OH)2 were in the original antacid?!
(0.344 g)
AP Chem Solution Stoichiometry: More practice
page 5
1. You Try it “Back titration” : An antacid tablet contains CaCO3 and other ingredients like starch
and sugar. The tablet is reacted with 10.0 mL of 3.00 M HCl. (Assume HCl is in excess and the
other tablet ingredients don’t react with HCl.) 5.00 mL of the excess HCl is then titrated with
exactly 8.00 mL of 1.25 M NaOH.
a. Write the balanced complete ionic equation for the reaction of CaCO3 + HCl.
b. How many mg of CaCO3 were in the tablet? !
(500 mg)
2a. Balance this redox equation: (Hint: some of the Cl’s oxidize, some don’t. Write HCl twice,
balance, then recombine for the mole ratio.)
!
KMnO4
+
HCl +
HCl
–––>
MnCl2 +
Cl2(g) +
KCl +
H2O!
!
b. How many liters of Cl2 gas are produced at 25.0 °C and a partial pressure of 0.90 atm when
50.0 ml of 0.450 M KMnO4 solution reacts with excess HCl? !
(a. 2, 16, 2, 5, 2, 8; b. 1.53 L)
!
c. 35.0 mL of 0.125 M HCl reacts with exactly 12.2 mL of KMnO4 solution as in the above
reaction. What is the molarity of the KMnO4 solution?!
(0.0448 M)
!
d. 125 mL of 0.270 M KMnO4 react with excess HCl. How many grams of MnCl2 are
produced? !
(4.25 g)
3. a. 15.2 mL of NaOH solution is titrated with 12.2 mL of 0.100 M Potassium hydrogen phthalate
(1 H). Then 13.2 mL of an H2SO4 solution is titrated with 24.0 mL of the NaOH solution. What
is the Molarity of the H2SO4?!
(0.0803 M NaOH, 0.0730 M H2SO4)
b. How many grams of Al(OH)3 would you need to neutralize 200 mL of the same !
! Molarity H2SO4 from part a?!
page 6
(0.759 g)
4. a. Balance this redox reaction of Nickel with Concentrated sulfuric acid. (Hint, the H2SO4 is
reduced and a spectator. After you balance it, combine the H2SO4’s for the mole ratio.)
!
Ni +
H2SO4 +
H2SO4 –––>
Ni2(SO4)3 +
H2S +
H2O
b. Some Nickel reacts with 50.0 mL of 12.0 M H2SO4 (which is in excess) and is allowed to react
overnight. 5.00 mL of the leftover H2SO4 is titrated with 12.7 mL of 3.00 M NaOH. How many
grams of Nickel reacted? !
(a. 8, 15, 4, 3, 12; b. 12.8 g Ni)
5. (This is a “back titration.”) Federal regulations set the upper limit of ammonia (NH3) in air in a
work environment of 38.5 µg NH3/Liter of air. To test the air quality in a manufacturing plant,
100.0 Liters of air is bubbled through 100 mL of 0.0105 M HCl. The reaction that occurs in the
acid is: NH3 + HCl –––> NH4Cl
!
10.0 mL of the reacted mixture is then titrated with exactly 13.1 mL of 0.00588 M NaOH (5.88 x
10–3 M NaOH). Is the manufacturer in compliance with regulations?
(47.6 µg/L of air, no)