 Okay, can we start? Okay. What is the answer of this question we were discussing? Yeah, what is the answer of this question proving? Why it is a? Why it is a tell me? See when you write down the expression for Kp for this reaction that will be the partial pressure of CO2 and Kp value is given which is 0.009 5 Okay, it's a partial pressure would be this Now it is saying that 1% CO2 in air is sufficient to prevent any loss in weight when M2 usually is heated, right? So 1% CO2 in air. So what is the partial pressure of CO2 in air? that will be equals to The pressure of 1 by 100 into the total pressure of air That will be 0.01 into P Or we can say 0.01 atmospheric one atmospheric pressure if you did Right, so now the question is 1% of CO2 is sufficient to prevent any loss In weight when M2 CO3 is heated Right Kp for this is given Question is how much would be partial pressure of CO2 have to be have to be to promote This reaction at 120 degree Celsius. Okay, so you see the partial pressure of CO2 for this reaction We are calculating which is 0.0095, which is almost equals to 0.01 Okay, so if the pressure in air and Pressure of CO2 from this reaction is almost same then the reaction will not proceed then nothing will happen in this reaction right but The question is how much would be the partial pressure of CO2 have to be to promote this reaction? Means what should be the partial pressure of CO2 so that the reaction will go in forward direction? So for that process what happens is the air also has this much Atmospheric pressure almost this much the partial pressure of CO2 that we get from this reaction must be less than 0.0095 Then only the reaction will proceed in forward direction So option is correct here question number nine. Oh, this is again. I'll repeat see what I said The reaction that you have from that reaction According to the Kp value the partial pressure of CO2 that we get here is 0.0095 and the partial pressure of carbon dioxide in air according to the question It is again one percent rights one percent of the pressure Right, so one by hundred into peer pressure of air, which is one atmospheric pressure So the partial pressure of CO2 in air will be 0.01 atmospheric So this 0.01 atmospheric is almost equals to the partial pressure of CO2 that we get from the reaction Which is given in the question Okay, so that is why that is why since the partial pressure is almost same So the reaction will not go in forward direction Not to proceed the reaction in forward direction the partial pressure of CO2 must be less than this value Then only it will go in forward direction. That's why option A is correct Okay, so this one is the simplest one option D is correct simple one again Kp is equals to Kcrt to the power del n Option D is correct here. What about the 11th one? Is it Kp greater than Kc because since del Ng is greater than 0 So Kp is greater than Kc 12 and 13 These are the question based on Kp and Kcr relation. Tell me quickly. 12th one is a correct 13th, it's not same You see here. It's Kc by Kp and here we have Kb by Kc So option A is correct not B. That is how you make mistakes in the exam also Yesterday also I was Discussing this and read out the question properly see it properly like what is the question? Correct. So minus one again This question You should get plus four right you actually have lost five marks That's why I'm saying don't make these kind of mistakes Okay, read out the question properly. Don't assume the question like okay. This is the question the question should be this See it properly. So Kc by Kp is option D. Next one 14 Equilibrium constant depends only on temperature. So it's remains same So I'm here and this week and poor week. What are you doing? Kc depends only on temperature, right? So with whatever you change the temperature the concentration that won't affect the equilibrium constant It would be same. These are the few key points. You should keep in mind. Don't Do these kind of mistakes? Question number 15. Is it be what week is getting be at the three question, right? And once we add the equilibrium constant will get multiplied Fine So when you add all these three The equipment will get this equation and for this the K will be equals to K1 into K2 into K3 So option B is correct. All of you are getting a 16th is a it's correct if you're getting so I'm not solving this okay 17. What about 17th all of you are getting C C is correct If you want me to solve, let me know done all of you So I'm here. What happened? I think he's not there. Anyways 18th and 19th I need equilibrium. Okay, I'll give you some questions on I think also solve these two 18th a is correct 19th you solve then I'll give you some question of any equilibrium 19th again, you are getting a this is also correct 18th a is right 19th a Is again, right? You want to solve some question of any equilibrium? Go on any equilibrium. I'll give you that question only then tell me yes or no, okay so You see this question. I'm giving you this question because I don't have the picture of this You have to find out the pH of pH of 10 to the power minus 4 molar OH minus solution 330 Kelvin if KW is given at this temperature and That will be equals to 10 to the power minus 13 0.6 Tell me the value of pH There are two options nine and nine point six There are two options. I'm giving you nine and nine points is other two are four and ten Tell me the answer nine point six exactly you are getting right. We know this KW is equals to what? Concentration of H plus into OH minus Which minus concentration is given so H plus concentration will be 10 to the power minus 9.6. So pH will be 9.6 Okay, next one you see What would be the pH of a solution obtaining by a mixing? We have to find out pH right. I'll just give you the question in short pH we have to find out and we have a mixture of five gram of of Acetic acid plus 7.5 Grams sodium acetate the volume is equals to 500 ml a a is given 1.75 into 10 to the power minus 5 pk a is 4.76 solve this 4.80 4.80 is exactly correct your concentration of CS3 COH Five divided by moral master. This is what? 60 into thousand by 500 so that will be 0.166 Concentration of salt if you calculate this will be 7.5 is the mass given and for this the masses 82 molar mass thousand by 500 It is 0.183 Okay, now this mixture gives you an acidic buffer right acidic buffer so pH for acidic buffer will be what pk a plus log of Concentration of salt divided by concentration of acid Okay, pH is 4.76 plus you will get some value over here the point is what here When you solve this you'll get log off 1.1 approximately Okay, now when you see the option like the option is given pH is 4.70 option a option B pH is less than 4.70 option C H of the solution will be equals to the pH of the acetic acid is 4.76 to 5 Like this. Okay. So you see if you are getting log greater than one over here Right greater than zero actually then this value will be what? Greater than 4.76 hence the option will be answer will be this if you exactly solve this your answer you'll get 4.80 Okay, so like this if log value is not given according to the option you can choose whether it is less than pk a value or More than pk a value any option Right, so this is the answer we have here Understood Can we move on next one you see you have to find out again pH of a buffer solution it is mentioned of a buffer solution prepared by dissolving 30 gram Of na2 co3 and for this we have 500 ml plus Aqua solution of this 500 ml of aqua solution of this and it is dissolved into one molar at cn 150 ml pH you have to find out for this a is given for co3 minus 5.63 into 10 to the power minus 11 what happened option you want options are 10.197 if you need any log value you can refer to log table or Calculated anything if you have 9.858 and the last one is 6.400 na2 co3 plus at cl forms nacl Plus na at co3. What is the molecular mass of this? What is the molecular mass of this 106 right? So number of moles for this will be what? 30 divided by 106 Is the number of moles and for this the number of moles will be what? It is given One or molarity So one into 151 50 Is the number of millimoles actually? and here the number of millimoles if you calculate for this so That will be 30 into thousand divided by 106 and when you solve this You will get 9.98 you are getting So you see here we have 283 283 millimoles and 150 millimoles right So when the reaction proceeds, what is the millimoles of this left 283 minus 150? Which is 133 Of this salt will left this will be zero this forms 130. Sorry 150 moles millimoles of this and 150 millimoles of this The solution containing any two co3 133 mole of any two co3 and At co3 minus acts as an acid buffer So ph formula will be pka plus log of salt salt is what co3 2 minus By acid is excuse me na h so3 or h co3 2 minus All these value you have to substitute Ph is nothing but minus log of k a plus log of co3 2 minus is 133 divided by 150 So this is what you have to solve and when you substitute all the value You will get 10.197 approximately 10.197 yes 133 133 By 150 because 150 millimoles of this na h co3 is forming And what is the salt left the salt left is this 133 to log of Concentration of salt by acid Understood tell me fast two more question. We'll discuss the last two questions Write down solid full twice gradually dissolved 10 to the power minus 4 molar and a 2 co3 solution At which concentration At which concentration of ba plus 2 precipitate of ba co3 begins to form begins to form ksp 2 3 is given And that is 5.1 into 10 to the power minus 9 What is the answer You want options 5.1 into 10 to the power minus 5 7.1 into 10 to the power minus 5 4.1 into 10 to the power minus 5 8.1 into 10 to the power minus 7 minus 2 You are getting a 5.1. It's correct What is the condition for precipitation The Concentration the product of ba plus 2 and co3 minus 2 Should be greater equal to its ksp value Then only the precipitation takes place Right so ba 2 plus concentration we have to find out So that will be what ksp 5.1 into 10 to the power minus 5 minus 9 divided by 10 to the power minus 4 So it will be 5.1 10 to the power minus 5 Option A is correct This is the condition Now the last question Today Acid is given HA plus plus A minus of 1 molar solution 5 Find Its Dissociation constant Condomia is getting 10 to the power minus 10 Nothing is given here So we assume the initial concentration is 1 0 0 Ph is given so we have the h plus concentration which is 10 to the power minus 5 And since 1 1 mole ratio we have So the concentration of h plus Would be equals to the concentration of A minus And how do we get this when this Acid will dissociate So when this 1 moles gives 1 To get 10 to the power minus 5 mole 10 to the power 5 moles of this must Dissociate So 1 minus 10 to the power minus 5 This is the equilibrium concentration of this So we can write down the K for this acid Is equals to h plus into A minus 10 to the power minus 5 Into 10 to the power minus 5 Divided by 1 minus 10 to the power minus 5 So since 1 is far Greater than 10 to the power minus 5 So we can neglect 10 to the power minus 5 with respect to 1 This is equals to 1 And hence K is equals to 10 to the power minus 10 approximately How did you get 10 to the power minus 3? What did you do? Sai Me Ram Chiran Don't do all this kind of mistake You are not losing 1 1 marks But you are losing 5 marks So whenever you get some easy questions Take your time and solve it Read the question properly And one more thing You are not going to solve all the questions So up the question And then read it properly And then solve These are the easy questions Don't lose marks in these kind of questions Anyways, so We will wind up the class here only And Next week probably I don't know, next week probably you will have class This week you don't have any chemistry class Tomorrow you don't have any chemistry class So wherever you feel Like you have less confidence Or something We just revise those topics I have sent you yesterday The Boron family questions Assignment I have sent you yesterday So today also I will send you Carbon and other groups Questions Those you have to go through Okay Bye bye See you soon