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REACTIONS IN SOLUTION Reactions are carried out in solution so the reactant particles can mingle freely and the

reaction can take place rapidly. In general, in a solution the solvent is the substance whose physical state doesn't change. If two liquids are mixed, the one present in the larger amount

is the solvent. When one of the components of a solution is water then H 2 O is taken to be

the solvent. All other components of the solution are solutes. Learn the definitions of concentrated and dilute.

The proportion of solute to solvent is specified by giving the solution's concentration. Molar concentration or molarity is a convenient concentration unit for measuring small

amounts of solute that are dissolved in a solution. The units of molarity is mol L-1 or it is also represented by M:

molarity =

moles of solute litre of solution

To calculate molarity you need to know the number of moles of the solute and the total

volume of the solution in which it is dissolved.. If you know the molarity of a solution, you can use it to calculate the number of moles of solute in a specified volume of the solution, or

to calculate the volume of solution needed to contain a specified number of moles of solute. Think of molarity as a conversion factor that relates moles of solute to volume of solution. You should be able to translate a label such as 2.50 M H 2 SO 4 into these factors: 2.50 mol of H 2 SO 4 in 1.00 L solution

1.00 L solution contains 2.50 mol H 2 SO 4

They can also be written with the volume units in milliliters. 2.50 mol H 2 SO 4 in 1000 mL solution

1000 mL solution contains 2.50 mol H 2 SO 4

In preparing a solution of a desired molarity, remember that the solvent (water. for example) is added to the solute until the desired final volume of the solution is reached. To

prepare 500 mL of solution, for example, we do not just add 500 mL of water to the solute.

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First the solute is dissolved in a small amount of water (usually in a volumetric flask) and then more water is added until the final volume is 500 mL NOTE: Terminology

Solution : a homogeneous mixture of two or more substances

A solution is made up of a solute (usually the substance in smaller amount) and solvent (usually the susstance in larger amount)

Concentration refers to number of solute particles per given amount of solution (usually volume or mass)

Methods for expressing concentration of solutions There are many ways of expressing concentration of solutions e.g. 1.

molarity

2.

mass percent

5.

mole fraction

6.

ppm (parts per million)

3.

molality

4.

normality

Only molarity and mass percent will be considered in this course.

Example 1: A student prepared a solution of NaCl by dissolving 1.461g of NaCl in a 250 mL volumetric flask. What is the molarity of the solution?

MM : NaCl = 58.44 g mol-1.

From the definition of molarity

M =

n V

Expressing the mole in terms of grams and molar mass(MM) and the volume in terms of litres the equation becomes =

=

g NaCl 1 x MM V

1.461g 1 x 58.44g mol −1 0.250L

3

=

0.1000 mol L-1 NaCl solution

Example 2: How many mL of 0.250 mol L-1 NaCl solution must be measured

to obtain 0.100 mol of NaCl

M=

V=

=

n NaCl V nNaCl M

0.100 mol 0.250 mol L−1

= 0.400L

convert to mL - x 1000

= 0.400L x 1000 mL .L-1 EXERCISE 1 1.1

= 400mL of NaCl soln. Is required.

Calculate the molarities of the following solutions:

(a) 0.350 mol NaHCO 3 in 0.400 L of solution (b) 0.250 mol KCl in 200 mL of solution

(c) 15.6 9 Of MgCl 2 in 300 mL of solution

(d) 1.85 g of AgNO 3 in 75.0 mL of solution

1.2 1.3

Calculate the number of moles of CaCl 2 in:

(a) 1.15 L of 0.840 M CaCl 2 solution. (b) 325 mL of 0 150 M CaC1 2

solution

What volume (in mL) of 3.00 M NH 3 solution contains (a) 1.35 mol of NH 3 ?

(b) 21.4 g of NH 3 ?

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1.4

How many grams of KNO 3 are needed to prepare 750 mL of 0.200 M KNO 3 solution?

EXERCISE 2.

2.1 Calculate the molarity of the solute in each of the following solutions (a) 1.14 mol KI in 1.50 L of solution

(b) 0.240 mol CaC1 2 in 500 mL of solution (c) 3.50 g of NaCl in 0.0500 L of solution (d) 4.25 9 MgSO 4 in 75.0 mL of solution

2.2 How many moles of KClO 3 are in 500 mL of 0.150 M solution?

2.3 How many moles of urea are in 250 mL of urine if the urea concentration is 0.320 M?

2.4 A normal adult excretes about 1500 mL of urine per day. If the urea concentration is 0.320 M, how many grams of urea are excreted per day?

2.5

Urea has the formula, CO(NH 2 ) 2

What is the molar concentration of each ion in the following solutions?

(a) 0.300 M AICl 3

Na 3 PO 4 2.6

(b) 0. 150 M (NH 4 ) 2 Cr 2 O 7

(d) 0.400 M Cr 2 (SO 4 ) 3

(c)

0.200

M

(e) 0 100 M Ba(OH) 2

Calculate the number of mL of 0.250 M CaC1 2 required to react

completely with 300 mL of 0.150 M AgNO 3 according to the equation, Ag+ + Cl- -+ AgCl(s)

2.7

Calculate the number of moles of solid AgCl that forms if 300 mL of

2.8

Ag+ + Cl- → AgCI(s) In Question 13.7 what will be the molar concentrations of the ions

0.240 M AgNO 3 is added to 200 mL of 0.480 M HCl. The reaction is: remaining in solution after the reaction is complete?

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2.9

In an experiment a student mixed 300 mL of 0.200 M Ba(OH) 2 with 500 mL of 0.100 M CuSO 4

(a) Write net ionic equations for any chemical reactions in this mixture.

that

occur

(b) For any solids formed, calculate their amounts in moles.

(c) Calculate the molar concentrations of any ions that remain solution after reaction is complete.

in

2.10 A 6.73g sample of Na 2 CO 3 is dissolved in enough water to make 250 mL of solution. (a)

(b)

What is the molarity of the sodium carbonate?

What are the molar concentrations of Na+ and CO 3 2- ions?

Ans. 0.254 M Na 2 CO 3 and CO 3 2- 0.508M Na+

2.11 What is the mass in grams of solute in 250 mL a 0.0125 mol L-1 solution of KMnO 4 ?

Ans. 0.494 g

2.12 What volume of 0.123mol L-1 NaOH, in mL, contains 25g of NaOH?

Ans.5080 mL

Remember the following terms Precipitate: A solid that is formed in a solution usually as the result of a chemical reaction. Solute. A substance dissolved in a solvent.

Solvent. Generally the substance in a solution that is present in largest amount. If one substance is a liquid it is normally considered to be the solvent.

Concentrated: A large proportion of solute to solvent in a solution.

Dilute: Very little solute dissolved in a solution-a low ratio of solute to solvent.

Concentration: A quantitative statement of the proportion of solute to solvent or of solute to the total amount of solution.

Molar concentration (molarity): A ratio of moles of solute to liters of solution. It is the number of moles of solute per liter of solution.

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Molar: A term that describes the molar concentration of a solute in a solution. It means moles of solute per liter of solution.

PREPARING SOLUTIONS BY DILUTION

You must know how to perform calculations that are necessary when dilute solutions are to be prepared from concentrated solutions.

When ever H 2 O is added to a solution it becomes diluted i.e. the molarity (or

mass percent) decreases. The amount of solute however remains in a solution remains constant as the solution is diluted, we can use the simple relationship Moles of solute before dilution = Moles of solute after dilution

nbefore

=

nafter

From the definition of molarity i.e. M =

n V

Therefore moles = M x V(L)

And

M j V j = M f V f = n which is constant

i and f refer to the initial and final solution.

It is important to note that mixing two solutions containing different ions will also involve dilution

e.g. mixing 1L of a 1mol L-1 NaCl with 1L of a 1 mol L-1 KBr will result in the concentration of all the ions decreasing V T = 1L + 1L = 2L

n 1mol = e.g. M NaCl = V 2L = 0.5mol L-1

Notice that in this case the concentration of NaCl decreased by half

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Examples 1.

How many mL of a 10.0 mol L-1 HCl solution is required to prepare 25.0L of 0.500 mol L-1 HCl solution?

n HCl (Before) = n HCl (After)

M HCl (B) x V HCl (B) = M HCl (A) x V HCl (A) V HCl (B)

Example 2

=

= =

0.500mol L−1 x 25.0L 10.0mol L − 1

1,25L

1250mL

If 82.5 mL of 6.25 mol L-1 solution is diluted to a final volume of 250 mL, What

is the molarity of the dilute HCl solution? n HCl (Before) = n HCl (After)

M HCl (Before) x V HCl (Before) = M HCl (After) x V HCl (After) M HCl (After) =

MHCl (Before) x VHCl (Before) VHCl (after )

6.25 mol L−1 x 0.0825 L = 0.250 L

= 2.06 mol L-1

You must know how to prepare a dilute solution from a concentrated solution. Finally, for safety reasons when diluting concentrated laboratory reagents with water. always add the concentrated reagent to the water.

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EXERCISE 3 3.1

3.2 3.3 3.4

How many mL of 3.00 M HCI must be used to prepare 500 mL of 0.100 M HCI?

How much water must be added to 50.0 mL of 1.00 M NaOH to produce 0.100 M NaOH?

If 4.00 mL of 0.0250 mol L-1 CuSO 4 is diluted to 10.00 mL with pure water what is the molarity of the copper salt in the solution?

0.00100 M

If 300 mL of H 2 O is added to 600 mL of 0.960 M H 2 SO 4 . what is the final molarity of the H 2 SO 4 ?

Note the following: Reagent. A term often used to refer to common chemicals that stocked in the laboratory. Dilution. The act of making a solution less concentrated by the addition of more solvent.

Volumetric flask. A special flask designed to hold a specified volume of solution when filled to the line etched around the neck of the flask.

ANALYSIS

The Stoichiometry of Reactions in Solution using molarity as the concentration unit The principal way "solution stoichiometry" problems differ from others that you've done so

far is that amounts of reactants and products can be specified as volumes of solutions of known concentrations. It is useful to remember that the product of molarity and volume gives moles: it is just necessary to be careful about the volume units.

molarity =

In terms of units

moles of solute litre of solution

Moles =volume x molarity Mol = L X mol L-1

Mol(unit)

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Example 1:

Copper is dissolved in dilute HNO 3 solution by the following reaction

3Cu + 8HNO 3 → 3Cu(NO 3 ) 2 + 2NO + 4H 2 O

How many mL of 3.00 mol L-1 HNO 3 solution can react with 10 g of Cu? Molar mass of Cu = 635 g mol-1

Calculate moles of Cu =

10g

63.5 g mol −1

= 0.157 mol Cu

Calculate moles of HNO 3

n HNO3 n Cu = From the equation 8 3 =

8 x 0.157 3

= n Cu x

8 3

mol of HNO 3

= 0.420 mol of HNO 3

Volume of HNO 3 is calculated from

M=

n therefore V = M

n V

0.420mol = 3.00mol L−1 = 0.14 L = 140 mL.

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Example 2 How many milliliters of 0.200 M Ba(NO 3 ) 2 are needed to react completely with 75.0 mL of 0.150 M Fe 2 (SO 4 ) 3 solution?

The net ionic equation for the reaction is

Ba2+ + SO 4 2- → BaSO 4 (s)

First determine how many moles of sulfate ion there are in the Fe 2 (SO 4 ) 3

solution. Then we can calculate how many moles of barium ion are needed, and finally we can calculate the volume of the Ba(NO 3 ) 2 solution needed.

The molar concentration of SO 4 2- in the Fe 2 (SO 4 ) 3 solution is SO 4 2concentration = 3 x (0. 150 M) = 0.450 M

Therefore, in 75.0 mL of this solution there is:

0.0750 L x 0.450 mol L-1 = 0.0338 mol SO 4 2-

From the stoichiometry of the net ionic equation, it is obvious that the number

of mole of Ba2+ needed for the reaction is also 0.0338 mol. Moles of Ba2+ = Moles of SO 4 2- = 0.0338 mol

The volume of the Ba(NO 3 ) 2 solution required is obtained from the molarity of

the solution. In this solution the Ba2+ concentration is 0.200 M because each

formula unit of Ba(NO 3 ) 2 contains one Ba2+. Therefore,

V(Ba2+ ) =

0.0338 mol 0.200 mol.L−1

V(Ba 2+ ) = =

n M

0.169 L

Convert litres to milliliters by multiplying by 1000

= 0.169 L x 1000 mL L-1 = 169 mL

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Note: limiting reactant calculations applies to solution chemistry as well. EXERCISE 4 4.1

Consider the following reaction that takes place in solution

CaCl 2 (aq) + 2AgNO 3 (aq) → Ca(N0 3 ) 2 (aq) + 2AgCl(s)

Suppose that we began with 200 mL of a 0.200 M solution of CaCl 2 (a)

(b) (c) 4.2

How many moles of AgNO 3 would be required to react completely with this amount of CaCl 2 ?

How many milliliters of 0.500 M AgNO 3 solution would be needed to contain this number of moles of AgNO 3 ?

Consider the reaction described in Question 13 above. Suppose that 200 mL of 0.150 M CaCl 2 solution are mixed with 180 mL of 0.220 M AgNO 3 solution. (a)

How many moles of CaCl 2 are in the first solution?

(c)

When the solutions are mixed, which is the limiting reactant?

(b) (d) (e) (f) 4.3

How many moles of CaCl 2 are there in this solution?

How many moles of AgNO 3 are in the second solution? How many moles of AgCI are formed in the reaction?

How many moles of the reactant in excess are left over after the reaction has stopped?

What is the molar concentration of the excess reactant in the reaction mixture after the reaction has stopped?

How many grams of Na 2 CO 3 are required for complete reaction with 25.0 mL of 0.155 mol L-1 HNO 3 . The equation for the reaction is Na 3 CO 3 + HNO 3 → NaNO 3 + CO 2 + H 2 O

4.4.

0.205g

What mass of Na must react with 125 mL of water to produce a solution that is

0.250 mol L-1 NaOH? ( Hint : write a balanced equation for the reaction between sodium and water).

0.718g

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4.5

A 0.2048 – gram sample of oxalic acid, H 2 C 2 O 4 , requires 24,87 mL of a particular NaOH(aq) to complete the following reaction. What is the molarity of the NaOH(aq)? The equation for the reaction is

H 2 C 2 O 4 (s) + 2NaOH(aq) → Na 2 C 2 O 4 (aq) + 2H 2 O

4.6

A 1.000-gram sample of an iron ore containing Fe 2 O 3 was dissolved in acid and all

the iron converted to Fe2+. The solution was titrated with 90,4 mL of 0.02000 mol L-

1

4.7

KMnO 4 . The reaction that occurs is as follows:

5Fe2+ + MnO 4 - 8H+ → 5Fe3+ + Mn2+ + 4H 2 O 72%

Calculate the % Fe in the ore.

A 25,00 mL sample of HCl was added to a sample of CaCO 3 . All the CaCO 3 reacted, leaving some some HCl(aq) over.

CaCO 3 (s) + 2HCl → CaCl 2 (aq) + H 2 O + CO 2 (g)

The excess HCl (aq) required 45,76 mL of 0,01221 mol L-1 Ba(OH) 2 (aq) to complete the following reaction.

2 HCl(aq) + Ba(OH) 2 (aq) → BaCl 2 (aq) + 2H 2 O

What is the mass, in grams, of the original CaCO 3 sample?

1.508 g

CHEMICAL ANALYSIS

Your study of stoichiometry so far should have convinced you that

(a)

if you know the balanced equation for a reaction occurring between two reactants,

(c)

and if you know the exact quantity of one of the reactants, then you can always

(b)

if the reaction is rapid and complete,

obtain the exact amount of any other of the substances in the reaction.

This is the essence of any technique of quantitative chemical analysis, the determination of the amount of a given constituent in a mixture.

Suppose, for example, you want to analyze a type of common clover for the quantity of oxalic acid, H 2 C 2 O 4 , in the leaves. We know this acid reacts with the base sodium hydroxide in aqueous solution according to the balanced equation

H 2 C 2 O 4 , (aq) + 2 NaOH(aq) →Na 2 C 2 O 4 (aq) + 2 H 2 O(l)

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You can tell exactly how much oxalic acid is present in a given mass of clover leaves if the following conditions are met:

1. the reaction is done in such a way that you know when the sodium hydroxide being added is exactly the amount required to react with all the oxalic acid present in solution;

2. the exact volume of the base is known when you have reached the point where the exact stoichiometric reaction has occurred; and

3. the concentration of the sodium hydroxide is known exactly.

These conditions are fulfilled in a titration, The solution containing oxalic acid is placed in

a flask along with a highly coloured dye, the purpose of which is explained below. Sodium

hydroxide of exactly known concentration is placed in a burette, a type of measuring

cylinder most commonly of 50.0 mL volume and calibrated in 0. 1 mL divisions. As the sodium hydroxide is added slowly to the acid solution in the flask, the acid is consumed by reaction with the base according to the net ionic equation H 3 O+(aq) + OH-(aq)

2 H 2 O(l)

As long as H 3 O+ ' from the acid is present in solution, every mole of OH- supplied by the

base from the burette is consumed by the H 3 O+. The point at which the number of moles of

OH- added is equal to the number of moles of H 3 O+ supplied by the acid is called the equivalence point. To indicate to us when this point has been reached, we added to the

solution an acid-base indicator, the dye mentioned above. The dye is selected for its

sensitivity to the H 3 O+ concentration in solution; it undergoes a strong colour change at a hydronium ion concentration as near the equivalence point as possible.

When the equivalence point has been estimated in a titration, the volume of base used from the beginning of the titration can be determined by reading the calibrated burette. If you know the concentration of the base in units of moles/litre, you can then calculate the exact number of moles of base used from:

Moles of base used = molarity of base (mol/L) x volume of base (L) Since the balanced equation for the acid-base reaction gives you the stoichiometric factor, you can use this mole-ratio conversion factor to convert moles of base added to the exact number of moles of acid present in the original sample.

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ACID-BASE TITRATIONS Suppose you have 1.034 g of clover leaves and you extract the oxalic acid from them into a small amount of water. This solution of oxalic acid is found to require 34.47 mL of 0.100 M

NaOH for titration to the equivalence point. What is the percentage by mass of oxalic acid in the leaves?

Solution Step 1.

Write the balanced equation.

Step 2.

Write an expression to show the reacting ration of acid to base in order to

H 2 C 2 O 4 , (aq) + 2 NaOH(aq) → Na 2 C 2 O 4 (aq) + 2 H 2 O(l

calculate the amount of acid required to react with the given amount of base

n acid n base = 1 2

0.100 mol L−1 x 0.03447 L = 2 Step 4.

= 0.00172 mol of oxalic acid

Calculate the number of grams of acid in solution. G acid = n acid x MM acid

= 0.00172 mol x 90.04 g mol-1

Step 5.

= 0.155 g

Calculate the weight percentage of H 2 C 2O 4 , in 1.034 g of leaves.

Weight percentage

=

0.155 g of acid x 100 1.034 g of leaves

= 15 %

In the preceding example, the exact concentration of the base, NaOH, was known. The

procedure by which the exact concentration of any reagent is determined is called standardization, and there are two general approaches to this as illustrated by the next example.

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CONCENTRATION EXPRESSED AS MASS PERCENT Another way of expressing the concentration of a solution is a ratio os the mass of solute to mass of solution expressed as a pecentage.

Mass percent thus refers to the mass of solute per 100g of solution. Notation : eg 40 % NaOH (m/m) (m/m) = mass by mass i.e.

40g of NaOH per 100g of solution

Example 1: A solution is prepared by dissolving 10g of NaOH in 150g of H 2 O. Express the concentration of the solution in terms of mass percent.

All that is required is the ratio of the mass of solute to mass of solution multiplied by 100

g solute x 100 Mass % (soln) = g solution =

10 g x 100 (10 g + 150 g )

= 6.25% (m/m)

6.25% is the amount of solute in the solution.

Example 1 : How many grams of NaOH are present in 300 g of 40% (m/m)

NaOH solution

Mass % (soln)

=

g (solute)

=

g solute x 100 g so ln

40 x 300g 100

=

120 g

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PRACTICE EXERCISE 1 :.What volume of concentrated H 2 SO 4 solution, which

is 96 % H 2 SO 4 by mass and has a density of 1.84g mL-1 is required to prepare 400mL of a 3.00 M H 2 SO 4 solution. Assume 100g of solution

∴

96 x 100g = 96.0g of H 2 SO 4 (solute) 100 g n( H 2 SO4 ) = MM

n( H SO4 ) = =

96,0 g 98.0 g mol −1

0.980 mol

Volume of 100g of solution

100 g −1 V= ρ = 1.84 g `mL = 54.3 mL = 0.0543 L g

M H2SO4

=

=

=

n V

0.980mol 0.0543L

18.0 mol L-1

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PRACTICE EXERCISE 2. Add in more examples of percent composition 24.

Concentrated HCl is 36.0% HCl by mass and has a density of 1.18 g mL. a)

What is the molarity of the HCl?

b)

What volume of the concentrated acid is required to produce 15.0 L 0.346 mol L-1

HCl?

11.60 M

445 mL