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Solution Stoichiometry

The concentration of a solution is the amount of solute present in a given quantity of solvent or solution.

M = molarity =moles of solute

liters of solution

M =moles of solute

liters of solution

Molarity (M)

Molality (m)

m =moles of solute

mass of solvent (kg)

Because density (volume) can change with temperature it is helpful to express solvent by mass when sample undergoes temperature changes

Includes solute volume

Excludes solute mass

No volumetric measurements needed; all mass

Measures of ConcentrationThe concentration of a solution is the amount of solute present in a given quantity of solvent or solution.

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M =moles of solute

liters of solution

(grams)

(mL)

L =moles of solute

Molaritymoles = Liters x Molarity

Calculate the Molarity and Molality of a H2SO4 solution containing 24.4 g of sulfuric acid in 198 g of water at 70°C to produce a 204 mL solution.

M =moles

L solution

M =0.249 mol

0.204 L

= 1.22 Molar H2SO4

Molarity/Molality Problem

24.4 g H2SO4

98.1 g H2SO4

1 mol H2SO4 = 0.249 mol H2SO4

m =moles

kg solvent

m =0.249 mol

0.198 kg

= 1.26 molal H2SO4

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What is the Molar concentration of the Potassium ion [K+] when 9.85 g of K2CO3 are dissolved in a 250 mL solution?

9.85 g K2CO3

138.2 g K2CO3

= 0.143 mol K+

M =0.143 mol

0.250 L= 0.57 M K+

Ion Molarity Problem

1 mol K2CO3

2 mol K+

K2CO3(s) → 2K+(aq) + CO3

-2(aq)

1 mol K2CO3

Preparing a Solution of Known Concentration from solids

Volumetric Flask Mix till dissolved Bring to desired volume

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What is the Molar concentration of the Sodium ion [Na+] when 23.4 g of NaCl and 34.1 g of Na2O are dissolved in 0.60 L H2O?

23.4 g NaCl 1 mol

58.5 g NaCl = 0.40 mol NaCl = 0.40 mol Na+

34.1 g Na2O 1 mol Na2O

62.8 g Na2O = 1.08 mol Na+

0.40 mol Na+ + 1.08 mol Na+ = 1.48 mol Na+

M =1.48 mol

0.60 L= 2.5 M Na+

(doubles from subscript)

Molarity Problem

1 mol Na2O

2 mol Na+

How many grams of potassium dichromate (K2Cr2O7) are required to prepare a 250-mL solution whose concentration is 2.16 M?

M =mol

Lmol = ML

Molarity Problem

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Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution (stock).

DilutionAdd Solvent

Moles of solutebefore dilution (1)

Moles of soluteafter dilution (2)=

M1V1 M2V2=Number of moles does not change

4.9 Dilution Practice

Describe how you would prepare 500 mL of a 1.75 M H2SO4 solution, starting with an 8.61 M stock solution of H2SO4.

Keep in mind that in dilution, the concentration of the solution decreases but the number of moles of the solute remains the same.

M1V1 = M2V2

4.9 Solution

Solution We prepare for the calculation by tabulating our data:

M1 = 8.61 M M2 = 1.75 M

V1 = ? V2 = 500 mL

Thus, we must dilute 102 mL of the 8.61 M H2SO4 solution with water to give a final volume of 500 mL

Describe how you would prepare 300 mL of a 0.4 M H3PO4 solution, starting with an 1.5 M stock solution of H3PO4.

Dilution Problem

M1V1 = M2V2

You have 250 mL of a 3.0 M Ba(OH)2 solution. What is the concentration if we add 150 mL of water to the solution?

Bell Ringer

M1 = 3.0 MV1 = 250 mL

M2 = ?V2 = 250 mL + 150 mL = 400 mL

Crash Course: Water and Solutions for Dirty Laundrywww.youtube.com/watch?v=AN4KifV12DA

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2) What is the Molar concentration of the Sodium ion [Na+] when 2.8 g of Na3PO4 and 4.5 g of Na2CO3 are dissolved in 85 mL?

1) How many grams of solid NaNO3 are needed to produce 125 mL of a 0.85 M NaNO3 solution?

3) How would you prepare 250 mL of a 0.65 M H2SO4 solution from a stock of 6.5 M H2SO4 solution?

4) You have 50 mL of 6.0 M NaF and 450 mL water are added, what is the new Molarity?

Bell Ringer

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TitrationsIn a titration, a solution of accurately known concentration is

added gradually to another solution of unknown concentration until the chemical reaction between the two solutions is complete.

Standard solution – solution with known concentration to be precisely added for comparison

Equivalence point – the point at which the reaction is complete

example) 1 mol H2SO4 2 mol NaOH(obtained from balanced equation)

TitrationsIndicator – substance that changes color at (or near) the equivalence point

Slowly add baseto unknown acid

UNTIL

the indicatorchanges color

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Titrations can be used in the analysis of:

Acid-base reactions

Redox reactions

H2SO4 + 2NaOH 2H2O + Na2SO4

5Fe2+ + MnO4- + 8H+

Mn2+ + 5Fe3+ + 4H2O

*Can involve color changes without indicator

1) Write Balanced equation for stoichiometry (mole-to-mole ratio)

Titration Steps

2) Determine moles of Standard solution used.(mol = M x L)

3) With stoichiometry, convert molesstandard to molesunknown

(train tracks)

4) Determine unknown Molarity using given volume (L)(M = mol/L)

Alternative Equation:MsVs = MuVu

Coefficient # Coefficient #

It takes 32 mL of 2.0M HCl standard to neutralize a 500. mL solution of Ba(OH)2. What is the concentration of Ba(OH)2?

Titration Problem #1

1) 1Ba(OH)2 + 2HCl → BaCl2 + 2H2O (Reacts 1:2)2) mol HCl = 2.0M x 0.032 L = 0.064 mol HCl

3) 0.064 mol HCl 1 mol Ba(OH)2

4) M Ba(OH)2 = 0.032 moles 0.500 L

= 0.064 M Ba(OH)2

2 mol HCl = 0.032 mol Ba(OH)2

MHVH = MOHVOH

Coefficient #Coefficient #

Alternative:

2.032 = MOH500

12MOH = 0.064

Titration Problem #2

How many milliliters (mL) of a 0.610 M NaOH solution are needed to neutralize 20.0 mL of a 0.245 M H2SO4 solution?

NaOH + H2SO4 H2O + Na2SO42 2

1) For every 2 moles base added, it neutralizes 1 mole acid

2) Next we calculate the number of moles of H2SO4 in a 20.0 mL solution:

3) From the Balanced Equation: 1 mol H2SO4 2 mol NaOH.

= 9.80 × 10-3 mol NaOH

4.9 x 10-3 mol H2SO4 2 mol NaOH 1 mol H2SO4

4) L = 9.80 x 10-3 mol 0.61 M NaOH

= 0.0161 L or 16.1 mL NaOH

Titration #2 Solution

Moles = M x L

0.245 M x 0.0200 L = 0.00490 mol H2SO4

Titration: Finding the Molar Mass of an Unknown

Lauric Acid is a short-chain fatty acid that is solid at room temperature and monoprotic. We dissolve 0.022 grams into 500. mL of water and then titrate it with 0.010 M NaOH.

If it takes 11.0 mL of NaOH to neutralize the fatty acid, what is the Molar Mass of Lauric Acid?

• Molar mass has the units grams/mole. We weighed out the mass of the solid acid in grams. Titration can tell us how many moles of acid are present in the same sample.

• Because it is monoprotic, it will react 1:1 with NaOH.

Moles H+ = ? Molar mass 0.022 grams

Molar Mass Titration Solution

1) Given information states it reacts 1:1

3) 1.1 x 10-4 mol NaOH 1 mol Lauric acid

4) Molar Mass = = 200 g/mol

1 mol NaOH= 1.1 x 10-4 mol

Lauric Acid

2) (0.010 M NaOH) x (0.0110 L) = 1.1 x 10-4 mol NaOH

1.1 x 10-4 moles 0.022 grams

Large component of coconut oil

~ 3-6% of milkC11H23COOH

Example: Redox Titration

A 16.42-mL volume of 0.1327 M KMnO4 solution is needed to oxidize 25.00 mL of a FeSO4 solution in an acidic medium.

What is the concentration of the FeSO4 solution in molarity?

The net ionic equation is

need to find

givenwant to

calculate

Redox Titration Solution

Solution The number of moles of KMnO4 (in 16.42 mL) = M x L

From the net ionic equation we see that 5 mol Fe2+ 1 mol MnO4-

M =moles of solute

liters of solution =

1.090 x 10-2 mol

0.025 L= 0.436 M FeSO4

(0.1327 M KMnO4) x (0.01642 L) = 2.179 x 10-3 mol

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Titration Bell Ringer

• 50 mL of Ba(OH)2 is measured out into an Erlenmeyer flask. The concentration is unknown.

• It takes 8.5 mL of 2.5 M H3PO4 standard to reach the equivalence point and neutralize the unknown base.

• Determine the concentration of Ba(OH)2 solution.

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Chemistry in Action: Metals from the Sea

CaCO3 (s) CaO (s) + CO2 (g)

Mg(OH)2 (s) + 2HCl (aq) MgCl2 (aq) + 2H2O (l)

CaO (s) + H2O (l) Ca2+ (aq) + 2OH- (aq)

Mg2+ (aq) + 2OH- (aq) Mg(OH)2 (s)

Mg2+ + 2e- Mg

2Cl- Cl2 + 2e-

MgCl2 (aq) Mg (s) + Cl2 (g)

1.3 g of Magnesium/Kg seawater

Many metals are found in the earth’s crust, but it is cheaper to “mine” from the sea

Precipitation

Slightly soluble

Precipitation

Electrolysis of MgCl2 (redox)

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Gravimetric Analysis Analytical technique based on the measurement of mass

• Precipitation: the analyte is precipitated out of solution by adding another reagent to make it insoluble. Then filtered and weighed.

• Volatilization: the analyte is converted to a gas and removed. The loss of mass from the starting material indicates the mass of gas.

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Gravimetric Analysis1. Dissolve unknown substance in water (if not already dissolved)

2. React unknown with precipitating reagent to form a solid precipitate

Reagent: chemical added to another substance to bring about a change

3. Filter, dry, and weigh precipitate.

4. Use chemical formula and mass of precipitate to determine amount of unknown analyte (chemical of interest in experiment)

A sample of an unknown soluble compound contains Ba+2 and is dissolved in water and treated with excess sodium phosphate.

If 0.411 g of Barium phosphate precipitates out of solution, what mass of Barium in found in the unknown compound?

3 x 137.3 g Ba+2 601.9 g Ba3(PO4)2

x 100% = 68.4% Ba in Ba3(PO4)2

0.684 x 0.411 g = 0.253 g Ba+2

Gravimetric Analysis Problem #1

We need to find 1st find the Mass % of the analyte in the precipitated compound using their respective molar masses

*It is not mandatory to convert to percentage form. The mass fraction can be used directly.

411.9 g Ba+2

601.9 g Ba3(PO4)2x 0.411g Ba3(PO4)2 = 0.253 g Ba+2

We have 250 mL of a Copper (Cu+1) solution. We add excess Na2CO3 to precipitate out 3.8 g of Cu2CO3. What is the [Cu+1] Molarity of the original solution?

Gravimetric Analysis Problem #2

1 mol Cu2CO3= 0.041 mole Cu+1

M =0.250 L

= 0.16 M Cu+1

Since we need to find moles of Cu+1 , it will be quicker to use train-tracks instead of % composition (either would work).

3.8 g Cu2CO3

187.0 g Cu2CO3

2 mol Cu+1

1 mol Cu2CO3

mol Cu+ 0.041

A 0.5662-g sample of an ionic compound containing chloride ions and an unknown metal is dissolved in water and treated with an excess of AgNO3.

If 1.0882 g of AgCl precipitate forms, what is the percent by mass of Cl in the original compound?

Gravimetric Analysis Problem #3

35.45 g Cl-

143.4 g AgClx 1.0882 g AgCl = 0.269 grams Cl

More Gravimetric Review Problems1) An unknown ionic compound contains Carbonate (CO3

-2).

To precipitate the carbonate, we add excess CaCl2 and collect 25.3 grams of CaCO3 precipitate. Calculate mass of Carbonate present in the original compound.

2) 50 mL of a solution contains an unknown amount of Ni+ ions. We add excess Na3PO4 to precipitate out 5.67 grams of Ni3PO4. What is the Molarity of Ni+?

3) 340 mL of an unknown solution contains the Silver ion (Ag+). When excess Na2S is added, 15.8 grams of precipitated Ag2S are formed. What is the Molarity of Ag+ in the solution?