Concentration of hcl m initial burette reading ml final...

Concentration of HCl: M Initial Burette Reading: mL Final Burette Reading: mL Volume (mL) HCl used: mL Volume (mL) of Ca(OH) 2: mL Concentration of Ca(OH) 2: M Concentration of HCl: M Initial Burette Reading: mL Final Burette Reading: mL Volume (mL) HCl used: mL Volume (mL) of Ca(OH) 2: mL Concentration of Ca(OH) 2: M Average Molarity of Ca(OH) 2: (1) Concentration of Ca 2+ (x): Concentration of OH - (2x): Show calculations: (2) What is the pH of this solution? (3) What is the experimental Ksp of Ca(OH)2? (4) Calculate G in kJ for the reaction at 25ᵒC given your experimental Ksp value. Explain why the sign of G supports the Ksp of calcium hydroxide using the magnitude of K and the sign of G to support your answer. (5) Calculate S in J at 25ᵒ for the reaction if H (heat of solution) is 132 kJ/mol K. Explain how the sign of S supports the dissolution of Ca(OH) 2 using the laws of entropy.

Transcript

00:01 Hello students today we are going to solve a question and we have to make a graph for the same.

00:05 So you can see what i have written already in this line that is your moles of hclu, that is concentration multiplied by volume.

00:13 And then we have modes of oh negative.

00:15 Oh negative is equal to h positive that is molar ratio.

00:20 Then we have oh negative wine insaturated solution, moles per volume.

00:23 Then we have ca2 positive iron insaturated solution and that is ca2 plus is equal to half of your oh negative ions.

00:32 Okay.

00:33 Then we have written in the end that is your solubility product of concentration of ca2 plus and oh2 ions in having coefficient two.

00:43 Okay.

00:44 So let's solve it one by one.

00:46 So i'm doing it personally for more of hcl use.

00:49 F.

00:50 0 .1 mole per litre multiplied by that is liter only right so we'll cut the little to liter and eventually we'll get answer 4 .1 multiplied by 10 yes to power minus 4 mole okay similarly for the second part point 1 mole per liter multiplied by 0 .005 to liter okay you will eventually get 5 .2 into 10 is to power minus 4 right then again point 1 mole per liter multiplied by 0 .0041, you will get 4 .1 into, sorry, multiplied by 10 is to power minus 4 mole.

01:45 Okay.

01:45 Here also write mole and more.

01:48 Okay.

01:49 Then for the last part, you'll see 0 .1 mole per liter, or divided by liter multiplied by 0 .0055.

02:01 D term okay you will end up getting 5 .5 multiplied by 10 is to the par minus 4 4 moles all right so we have solved moles of xcel use now next is moles of oh negative that would be 4 .1 multiplied by 10 is to per minus 4 same is the case as such as mentioned in the upper one table then 4 .1 multiplied by 10 is per minus 4 then we have 5 .5 multiplied by 10 is 4 minus 4 okay then oh it's saturated solution how you are going to calculate by 10x4 minus 4 divided by 0 .01.

02:41 You'll end up getting 0 .041 moles.

02:47 Okay similarly 5 .2 multiplied by 10x4 minus 4 divided by 0 .01 again you'll get 0 .052 mole then similarly you'll get over here.

03:02 0 .041 when you will divide it with 0 .01 when you will divide 4 .1 multiply by 10 is 4 by 0 .01 you'll get 0 .041 more and here you'll get 0 .055 more right then in the next case you will get you will have to divide the above provided values that is 0 .041 by 2 okay because calcium concentration is equal to half of your os negative concentration so you'll end up getting 0 .0205 mole here you'll end up getting 0 .026 mole here also 0 .0205 more here 0 .0275 mol okay then here you'll solve the solubility product of ca2 plus and oh negative which would be here it is going to be 3 .44 multiplicated by 10 is 4 minus 5 here it is going to be 7 .03 multiplied by 10 x to 5 then here it is going to be 3 .44 into 10 to 1 minus 5 i have solved that already you can do it again see this slide once again you'll understand that eventually what i have done over there so likewise i have calculated the solubility product as well but this is this question is not over here.

04:32 Let's see.

04:33 So now let's calculate the average value...

Question

Please give Ace some feedback