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Reactions performed at 298K, Kw = 1x10-14 mol2dm-6a) Calculate the pH of the solution formed when 25.00 cm3 of 0.200 mol dm-3 HCl is added to 20.00cm3 of 0.160 mol ...

Question

Reactions performed at 298K, Kw = 1x10-14 mol2dm-6a) Calculate the pH of the solution formed when 25.00 cm3 of 0.200 mol dm-3 HCl is added to 20.00cm3 of 0.160 mol dm-3 NaOH. b) Calculate the pH of the solution formed when 15.00cm3 of 0.300 mol dm-3 H2SO4 is added to 20.00cm3 of 0.800 mol dm-3 NaOHplease provide answer to both parts.

Reactions performed at 298K, Kw = 1x10-14 mol2dm-6 a) Calculate the pH of the solution formed when 25.00 cm3 of 0.200 mol dm-3 HCl is added to 20.00cm3 of 0.160 mol dm-3 NaOH. b) Calculate the pH of the solution formed when 15.00cm3 of 0.300 mol dm-3 H2SO4 is added to 20.00cm3 of 0.800 mol dm-3 NaOH please provide answer to both parts.



Answers

$\mathrm{A} 500 .$ -mL solution consists of 0.050 $\mathrm{mol}$ of solid $\mathrm{NaOH}$ and 0.13 $\mathrm{mol}$ of hypochlorous acid $\left(\mathrm{HClO} ; K_{\mathrm{a}}=3.0 \times 10^{-8}\right)$ dissolved in water.
(a) Aside from water, what is the concentration of each species that is present?
(b) What is the pH of the solution?
(c) What is the pH after adding 0.0050 mol of HCl to the flask?

So the standard free energy change for the reaction is given by the following relationship. Delta G not equal to the sum malls gives the energy of formation of the products. Subtract the exact same of the reactant. And so what we can do is serve our values into this equation to get positive 368.0 kg joules per mole. And then for the second part of this video here we're looking at a slightly different equation. Delta G is equal to delta jeannot at R T L n Q. Where Q is the reaction quotient what Q is equal to O H minus square Because it's so key metric coefficients ph two divided by B r minus squared. So we need to determine all of these three values to then plug them in. Ask you for this equation. Work you you should find equals 1.35 times 10 to the minus eight. Following on from this, you need to convert temperature to kelvin. So we have 25 degrees Celsius. We add 273.15 to get to 98.15 kelvin. Then we substitute all of our values into the second equation, including the queue that I've listed here for us all. And so we get delta G. That is 323 kg jewels.

We are asked to perform some calculations for a hypothetical weak acid H. A. Combining with sodium hydroxide. Were given that zero point 20 moles of H. A. And 0.80 moles of O. H. Are diluted to a volume of one leader. Giving us the concentrations of 0.20 and 0.80 respectively. For eight we're told that the ph is 4.80 Find P. K. A. Okay, so for PK eight um let's go ahead and write our equation here. Mhm. These are all areas except for the water. And we're told that our initial concentrations are 0.20 0.80 and zero. So we are going to decrease the concentration as follows and increase the concentration. Sorry, equilibrium concentrations will be 0.12 and 0.80 And if we take 10 to the minus 4.80 that will give you my H plus concentration. Which I got as I believe I got 1.585 times 10 to the minus 50. You know that the peak A. Will be equal to the H. Plus concentration times a minus concentration divided by H. J. We now have all of our values so we can substitute as follows. And what it did my calculations for part A. I got 1.6 times 10 to the minus fifth Pekka. Excuse me that wasn't my P. K. That was my K. A. Yeah and to find my Pekka we need to take the negative blog of 1.6 times 10 to the minus five. And for that I got 4.98 So that was part one Per cute. How many more moles have to be added to increase ph to 5.0 This one is actually pretty easy. We're going to take our same concentrations and we're going to call the amount of base we need to add B. And our new H plus concentration that we're going to be dealing with is 1.0 times 10 to the minus five. So we're going to simply set up are same calculation where R. P. K. Was one, I think it was 1.6 time. I gotta go look 1.6 times 10 to the minus five. And this time our new concentration will be 1.0 times 10 to the minus five. And we've got our initial concentration here was we're going to see how much do we have to add? How much be and remember that for every bit of B. We add we consume more of our acid. So we're gonna set our equation up like this multiply each side and I'll go ahead and write this out because I so bad at keeping track. And numbers 1.272 times 10 to the minus sixth minus 1.6 times 10 to the minus 50 B equals eight times 10 to the minus seventh, plus one times 10 to the minus 50 B. Now it's pretty easy to do. So I got 4.72 times 10 to the minus seventh equals 2.6 times 10 to the minus 50 be. And I got B equals two point 29 times 10 to the minus second moles. And this is the leader, so it'll be moles per liter same thing. Um And we can do to sig figs. So I got 2.3 times 10 to the minus second moles of any O. H. And we're done.

Problem 73 in this question. Give this compound. And we need to find out PS and Qs. And concentration for this reaction. So initial concentration of it's Tyler mine is given a question that is zero point 075. M. And initial concern personal. It's constitute iron is zero and zero. Now at equilibrium let X. M. O. Or it is from then X. And Of its iron will also form so remaining concentration of Patella mine at equilibrium is 0.0 75- X. No we write KB. For this reaction that is concentration all product that it's easy to mm hmm as high and as three positive into concentration of Oh it's negative divided by concentration of Ayatollah mine That it's c. two H 5 and S. Two. Now put the value access quiet. You hired by 0.0 75 -X. Now the value of access are very less as compared to 0.075. Now we can write access choir divided by 0.075 Equals to 6.4 into 10. To the power minus for value of cave is given the question. Now, resolving we get Value of X. That is equals two Concentration of Os. That is six nine into 10 to the power -3. a.m. Therefore we can write Yes equals 2 -6 hydrogen ion concentration. Sorry? Yes, he goes to 14 minus minus log hydrogen and concentration. Now we put the value that is putting the value of Haryana and concentration 14 minus minus log 6.9 into turn to the Power Ministry. After concluding that we get the value of Phds 11 point eight ft. So this is a P. S. Show loosen and this is a where's and concentration in the solution. So this is our final answer.


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