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Chemistry Exercises Assignment Sample

INTRODUCTION

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A. Acids and Bases.

  1. The relationship between pH and [H+] is 

pH=-log[H+]

Or, [H+] =10-pH

Here the pH of the solution is 3.5.

Therefore the [H+] =10-3.5M

The relationship between pOH and [OH-] is 

pOH=-log [OH-]

Or, [OH-]=10-pOH

Here the pH of the solution is 12.5.

So, here the pOH of the solution is (14-12.5) =1.5

Therefore the [OH-]=10-1.5M

  1. In 0.05M HCl solution the concentration of the acid is [H+] =0.05M

Therefore, the pH is=-log[H+] =-log0.05=1.3

In 0.65M NaOH solution the concentration of the base is [OH-]=0.65M

Therefore, the pOH is=-log [OH-]=-log0.65=0.2

Therefore, the pH is= 14-pOH=13.8

  1. Kw=[H+] [OH-]

Or, -log Kw= -log[H+]-log [OH-

Or, -log3.0 × 10–14= -log[H+]-log [OH-]

Or, 13.52= -log[H+]-log [OH-]

Or, 13.52=pH + pOH

Therefore, at 40 degrees Celsius, the neutral point of any solution is 13.52/2=6.76

So, the solution which shows a pH value of more than 6.76 becomes basic. Therefore, here the solution shows pH 7 at the same temperature where the Kw of water 3.0 × 10–14 is slightly basic.

For every chemical, pH is one of the most important variables to find out the basicity or the acidity of any solution (Orgill et al. 2019). Here, like the above process, the pH of any acidic and basic solution can be measured. The pH value is slightly dependent on the temperature. Now the pH of any very diluted solution can go near to the 7 as well as the pH of the water can be diverted from the exact value of pH 7 at room temperature. Therefore, it can be said from the above calculation that at 40 degrees celsius the solution shows pH 7 is slightly basic.

B. Titration.

Here the concentration of hydrochloric acid (Sa)=0.1M

Here the volume of HCl needed for the titration is =30ml, 25.5ml, 25.7ml and 25.6ml

As the volume 30ml is very much diverted from the other volume then that can be considered as the volume with man-made or technological error. Therefore, that value is avoided and the mean value from the other three values is considered as the value of the volume of HCl in the titration. 

So, here the volume of HCl needed for the titration is (Va)=25.6ml

The volume of sodium carbonate is (Vb)=25ml

Now, as here 2molar HCl is needed to neutralize 1 molar Na2CO3

Therefore, the concentration of the Na2CO3 is (Sb)=(Sa*Va)/Vb*2=0.1*25.6/25*2=0.0512M

Titration is one of the most common processes to determine the unknown strength of any solution. Hereby the help of an indicator that changes its color in the accurate change of the nature of a solution the unknown strength of any sample can be determined in this process from a known strength of a sample (Stull et al. 2018). 

C. Oxidation and Reduction.

  1. In CaCl2 oxidation number of Ca is +2 and Cl is -1

In PbO2 oxidation number of Pb is +4 and O is -2

In NaH oxidation number of Na is +1 and H is -1

In Na3PO4 oxidation number of Na is +1 and PO4 is -3

In O2 oxidation number of O is 0

The oxidation number is one of the major chemical properties which mainly indicates the overall numbers of election gaining or electron losing by any molecules after doing a particular reaction. Now any molecule can have a single oxidation number like Na which generally shows a +1-oxidation number in most of the reactions (Yang et al. 2021). Now from the example of the above compounds, it can be said that depending on the different molecules one atom can show two or more oxidation numbers like here oxygen shows -2 oxidation number in PbO2 while oxidation number in O2.

  1. Unbalanced ionic equation of the given reaction

Cr3+ (aq) + 2Br0 (aq)→ CrO42- (aq) + Br- (aq) in basic medium

Half reaction of oxidation: Cr3+ (aq) → Cr+6O42- (aq)

Half reaction of reduction: 2Br0 (aq)→ 2Br- (aq)

Other than the oxygen and the hydrogen atoms balance of the other atoms in the half-reaction are:

Cr3+ (aq) → Cr+6O42- (aq)

2Br0 (aq)→ 2Br- (aq)

Balance of the oxygen atoms in the half-reactions by the addition of the water molecules in the appropriate part is: 

Cr3+ (aq) + 4H2O (l)→ CrO42- (aq)

Balance of hydrogen atoms in the above half-reaction by adding the H+ ion in the appropriate side is:

Cr3+ (aq) + 4H2O (l)→ CrO42- (aq) + 8H+ (aq)

Now by the addition of H+ ion in the basic medium, the balance of the total hydrogen ion is:

Cr3+ (aq) + 4H2O (l) + 8OH- (aq)→ CrO42- (aq) + 8H+ (aq) + 8OH- (aq)

Or, Cr3+ (aq) + 4H2O (l) + 8OH- (aq)→ CrO42- (aq) + 8H2O (l)

Or, Cr3+ (aq) + 8OH- (aq)→ CrO42- (aq) + 4H2O (l)

Balance of the overall charge of both side of the both half reaction by the addition of electrons is:

Cr3+ (aq) + 8OH- (aq)→ CrO42- (aq) + 4H2O (l) + 3e

2Br0 (aq) + 2e- → 2Br- (aq)

The overall balanced reactions based on the total number of electors are:

2Cr3+ (aq) + 16OH- (aq)→ 2CrO42- (aq) + 8H2O (l) + 6e

6Br0 (aq) + 6e- → 6Br- (aq)

Overall balanced final ionic equation is:

2Cr3+ (aq) + 16OH- (aq) + 6Br0 (aq) → 2CrO42- (aq) + 8H2O (l) + 6Br- (aq)

Overall balanced final equation is: 2Cr (OH)3 + 3Br2 + 16OH- (aq) → 2CrO42- + 6Br- + 8H2O (l)

  1. Given ionic equation of the reaction:

COOH- (aq)+ MnO4- (aq)→ CO2 (g)+ Mn2+ (aq) in acidic medium

Half reaction of oxidation: C+3OOH- (aq) → C+4O2 (g)

Half reaction of reduction: Mn+7O4- (aq)→ Mn2+ (aq)

Balance of the atoms of the half-reactions other than the oxygen and the hydrogen atoms are:

COOH- (aq)→ CO2 (g)

MnO4- (aq)→ Mn2+ (aq)

Balance of the oxygen atoms in the half-reactions by the addition of the water molecules in the appropriate part is: 

MnO4- (aq)→ Mn2+ (aq) + 4H2O (l)

Now, the Balance of hydrogen atoms in the above half-reaction by adding the H+ ion in the appropriate side is:

COOH- (aq)→ CO2 (g) + H+ (aq)

MnO4- (aq) + 8H+ (aq)→ Mn2+ (aq) + 4H2O (l)

Balance of the charge by the addition of electrons in the appropriate side is:

COOH- (aq)→ CO2 (g) + H+ (aq) +2e-

MnO4- (aq) + 8H+ (aq) +5e-→ Mn2+ (aq) + 4H2O (l)

The equalised reactions based on the number of electors are:

5COOH- (aq)→ 5CO2 (g) + 5H+ (aq) +10e-

2MnO4- (aq) + 16H+ (aq) +10e-→ 2Mn2+ (aq) + 8H2O (l)

Overall balanced final ionic equation is:

2MnO4- (aq) + 16H+ (aq) + 5COOH- (aq)→ 2Mn2+ (aq) + 8H2O (l) + 5CO2 (g) + 5H+ (aq) 

Or, 2MnO4- (aq) + 11H+ (aq) + 5COOH- (aq)→ 2Mn2+ (aq) + 8H2O (l) + 5CO2 (g)

Or, 2MnO4- (aq) + 6H+ (aq) + 5HCOOH - (aq) → 2Mn2+ (aq) + 8H2O (l) + 5CO2 (g)

Overall balanced final equation is: 5 HCOOH + 2MnO4-+ 6H+ (aq) → 5CO2 + 2Mn2++ 8H2O (l)

There are various kinds of methods and processes in balancing a big chemical reaction. The half-reaction method is one of the most common and most used methods among the other methods (Tee et al. 2018). Here in the case of acidic medium, this half-reaction method is carried out by the addition of the H+ ion. While in the case of the basic medium reaction this process follows the balancing by the addition of the OH- ions.  In the above two examples, the balancing reaction in both different kinds of medium is given. In this way, the overall balancing of different kinds of reactions in different mediums can be achieved. 

D. Energy of reactions and Reaction kinetics

  • Balance equation of the combustion reaction is: CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l)

  • The combustion of methane is an exothermic reaction. The reaction which absorbs heat is called the endothermic reaction while the reaction which releases heat in nature is called the exothermic reaction. As here this reaction releases ΔHo = -890 kJ amounts of heat with every molar combustion of methane. Therefore, this reaction can be called an exothermic reaction (Widarti et al. 2017).

  • Produced energy from the combustion of 1 mole methane is= -890 kJ

Therefore, the energy which will be produced from the combustion of 2.7 moles of methane which they are let to burn with excess oxygen is= (-890 kJ/mole) *2.7mole=-2403 kJ

  • Now in the case of methane 1 mole= 16.04 grams of methane.

So, 15 grams of methane= (15/16.04) mole= 0.935 mole

So, produced energy from 15 grams of methane is= (-890 kJ/mole) *0.935mole=-832.15 kJ

  1. The steps of the nitric acid manufacturing reactions from the ammonia is as follows:

N2 (g) + O2 (g) → 2NO (g) 

2NO (g) + O2 (g) → 2NO2 (g)

N2 (g) + 2O2 (g) → 2NO2 (g) 

Now here the standard enthalpy of the first step is ΔHo = +180.6 kJ/mole and the last step is ΔHo = +66.4 kJ/mole.

Therefore, the standard enthalpy of the 2NO (g) + O2 (g) → 2NO2 (g) is: (+66.4- 180.6) kJ/mole= -114.2 kJ/mole

  1. C6H6 (g) + 3H2 (g) ? C6H12 (g) this reaction releases heat in the nature and called the exothermic reaction.

  • Equilibrium reaction is one of the most common reactions of chemistry. Here both sides of the reaction which is the reactant and the product stay in balanced equal amounts according to the reaction formula. In the case of the exothermic reaction as heat is produced in nature as the reaction happens (Bhukdee and Limpanuparb, 2020). So here the heat is considered as the product of the reaction. So, a shift towards the right side will be observed here.

  • In case of any equilibrium reaction when the size of the container is enlarged then the equilibrium here moves to the side in which side the volume is higher compared to the other side. Now in this case the reactant of the reaction has four molars of gas molecules while the right side of the reaction has only one molar of gas molecules. Therefore, here equilibrium will shift towards the left direction.

  • Similar to the above point in any equilibrium when the container is reduced or the pressure is increased by any other forces then reaction generally shifts to the side where the volume is lower compared to the other (Matanovic et al. 2017). So hereby decreasing the size of the container or by increasing the overall pressure by any method the reaction will be shifted to the right side of the equilibrium. 

  • In case of any equilibrium reaction when the volume of any side is increased then to maintain the equilibrium of both sides of the reaction moves to the adverse direction of the reaction of where the volume was increased. Now in this reaction when the H2 gas is put on the medium the overall magnitude of the reactant increases. This leads to the shifting of the equilibrium to the product side.

  • Now from the above point when the C6H12 gas has attached the equilibrium then the volume of the right side of the reaction increases (Al-Gaashani et al. 2019). So here for adding the C6H12 gas to balance the overall equilibrium the reaction shifts to the left side of the right side of the equilibrium.

Reference List

Journal

Al-Gaashani, R., Najjar, A., Zakaria, Y., Mansour, S. and Atieh, M.A., 2019. XPS and structural studies of high quality graphene oxide and reduced graphene oxide prepared by different chemical oxidation methods. Ceramics International45(11), pp.14439-14448.

Bhukdee, D. and Limpanuparb, T., 2020. Matching Five White Solids to Common Chemicals: A Dissolution Calorimetry and Acid–Base Titration Experiment. Journal of Chemical Education97(8), pp.2356-2361. 

Matanovic, I., Chung, H.T. and Kim, Y.S., 2017. Benzene adsorption: A significant inhibitor for the hydrogen oxidation reaction in alkaline conditions. The journal of physical chemistry letters8(19), pp.4918-4924.

Orgill, M., York, S. and MacKellar, J., 2019. Introduction to systems thinking for the chemistry education community. Journal of Chemical Education96(12), pp.2720-2729. 

Stull, A.T., Gainer, M.J. and Hegarty, M., 2018. Learning by enacting: The role of embodiment in chemistry education. Learning and Instruction55, pp.80-92. 

Tee, N.Y.K., Gan, H.S., Li, J., Cheong, B.H.P., Tan, H.Y., Liew, O.W. and Ng, T.W., 2018. Developing and demonstrating an augmented reality colorimetric titration tool. Journal of Chemical Education95(3), pp.393-399. 

Widarti, H.R., Permanasari, A. and Mulyani, S., 2017, February. Students’ misconceptions on titration. In Journal of Physics: Conference Series (Vol. 812, No. 1, p. 012016). IOP Publishing. 

Yang, F., Lai, V., Legard, K., Kozdras, S., Prieto, P.L., Grunert, S. and Hein, J.E., 2021. Augmented Titration Setup for Future Teaching Laboratories. Journal of Chemical Education98(3), pp.876-881. 

 

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