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6What is the pH of a 0.5 M ammonia solution? Ka (NH4+) = 5.6 x 10-10 (6 Puan)9.22.5none4.811.5...

Question

6What is the pH of a 0.5 M ammonia solution? Ka (NH4+) = 5.6 x 10-10 (6 Puan)9.22.5none4.811.5

6 What is the pH of a 0.5 M ammonia solution? Ka (NH4+) = 5.6 x 10-10 (6 Puan) 9.2 2.5 none 4.8 11.5



Answers

Find the pH of the substance with the given hydronium ion concentration. Ammonia, $2.5 \times 10^{-12}$

Everyone in this portion we have the salt in h four, see ill, which is This soul is considered ascetic, sold because it come from ammonia, which is a weak base which I have a value off baby and strong acid, which is HTC. So now we can count their concentration off. Each postive resulted from these acidic sort which will be the rules off the value off kw divided by the K off the base times by the concentration off the soul. So it will be equal the root off kw timber Negative 14. Dividing boy, the value off K B, which is one point the eight times my team more negative. Fine. All these times, boy, or going to four to So the concentration off each must It will be 1.5 to 75 turns. Boynton bore negative five muehler. And so the value off the actual before bowing to the 816 which ive around four point the A to thank you

So this question is asking us what is the Ph of a solution that consists of is your point to Mueller ammonia and 0.2 Moeller ammonium chloride. So here I have written just what's in the solution. We have point to, uh, molars of ammonia and then point to Mueller ammonium chloride. And so the way that we are going to calculate pH is we're gonna use the Henderson Hasselbach equation. So here I have two forms of it written one to calculate ph and one to calculate P O. H. Now in the book, we are given the PK or were given the KB of ammonia, which is 1.8 times tend in the neg. Fifth. And that is, um, what's called an ionization constant. So it just tells us, basically, how often is ammonia going to lionize into NH four plus, which is the congregate acid of ammonia? So the equation that we're gonna since we have K B, the equation that we're gonna use is P. O. H. And at the end of this, what we're gonna do is to get pH. We're going to essentially use this equation, which is P O. H. class P H equals 14. And you can either. This is the This is just the way I'm choosing to do it. You could just calculate ph straight away if you have to peek a of ammonia and you just have to be careful about which one that you are using as your country get acid at base. So we'll work with the equation I have in green there. So we'll do. The PDO age of Among of this solution is gonna be the p k B, which is just the negative log the negative log of our K V, which is 1.8 times 10 to the negative. Fifth. Like sad. We'll make that parentheses in green and then plus the log of the concentration of the conjugated acid, Which in our case, is this NH four plus I on. So we know that the concentration of our image four plus c l is going to be 40.2 moles. So that's what we're gonna put up here. So point to malls are muller divided by the concentration of and H three, which is 0.2 no. And so what we're gonna see is that the p. O age. If you use a calculator to work this out is p o age equals So the negative log of 1.8 times 10 to the negative fist. If we could work that out on a calculator, it's going to be four point seven five. Just about so 4.75 And then we're adding the log of essentially one, which is zero his 0.2 of a point to his lunch. And then, since we have mod function, that's 10 to so that's our P O. H. It's 4.75 and if we use this equation here, Teoh calculate P H, we get nine point, it's you sighs, and that is gonna be equal two ph for this solution.

To calculate the ph at the equivalence point we need to know the concentration of ammonium that is being created in the titillation. At the equivalence point. To do this, we'll assume all of the ammonia is converted into ammonium. So the moles of ammonia that we start with, which can be calculated by taking the volume and leaders multiplied by the polarity is now the moles of ammonium that we have. We then divide by the new volume, which will be 50 million liters because the concentrations are the same and the stock geometry is 1 to 1. We need equal volumes of the 2 25 plus 25 gives us 50 or 500.5 liters. This gives us a concentration at the equivalence point of .060 Mueller. We can solve for K. A. Of ammonium by taking KB of ammonia and dividing it into KW and we get 5.56 times 10 to the negative 10. This is small enough. We can make our assumption. So hydro knee um concentration will be equal to the square root of K. A, multiplied by the concentration of the acid, And we get 5.77 times 10 to the -6. We take the negative log of this to get the ph and we get 5.24 and an indicator that has 5.24 in its transition range is metal purple.

We have a reaction off carbon monoxide with chlorine to form carbonite chloride. The reaction is depicted as shown so carbon monoxide plus chlorine gives me see who steal too. We have value off. KP is equal to 3.10 at 700 kelvin temperature. We have the equilibrium constant K P is equal to pressure off C O seal, too. Divided by pressure off, CEO multiplied would seal, too. That is equal to 3.10 So using this initial partial pressure, we prepare an I C E table represented as this. We substitute the values in equilibrium constant and solve the quadratic equation. Tau obtain a value off X. We're next. Is the pressure off carbon monoxide react to reach an equilibrium, So solving the quadratic equation and determine the value off X. Let a solvent ex turn out to be X equal zero point 645 divided by two or the value off X will be 0.566 divided by two, which equals the value of X equal 0.3 to 2, or the value off excess 0.238 Here the value. 0.32 is discarded as it will give a negative value off partial pressure of carbon monoxide. So, bitches, 215 minus X, where X is equal to 0.283 This will be the value off X that will be used in for dissolving the question. Now we have the partial pressure at equilibrium as calculated, shown off each and every respective guests the pressure that equilibrium will be the some off the some off each gases we can clean. The more fraction off COC l that is carbon include right at equilibrium, which is more faction is you could do pressure off a particular gas divided by the total pressure substituting in the values we have. More fraction is equal to 0.879


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