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Monoprtic weak acid; HA, dissociates in water accordingHACaq) + H,O() = H,ot(aq) A-(aq)The equilibrium concentrations of the reaclants and products are (HA] 0.140 M...

Question

Monoprtic weak acid; HA, dissociates in water accordingHACaq) + H,O() = H,ot(aq) A-(aq)The equilibrium concentrations of the reaclants and products are (HA] 0.140 M, [#,ot] 40 * 10 ' M,and I4 | 40 * i0 M. Calculale Ihe K , Value for the acid HA,rcucuion

monoprtic weak acid; HA, dissociates in water according HACaq) + H,O() = H,ot(aq) A-(aq) The equilibrium concentrations of the reaclants and products are (HA] 0.140 M, [#,ot] 40 * 10 ' M,and I4 | 40 * i0 M. Calculale Ihe K , Value for the acid HA, rcucuion



Answers

The ionization constant of a very weak acid, HA, is $4.0 \times 10^{-9} .$ Calculate the equilibrium concentrations of $\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{A}^{-},$ and $\mathrm{HA}$ in a $0.040 \mathrm{M}$ solution of the acid.

The dissociation reaction can proceed. Us. A glass of water leads to Hi, Don't you're mine Plus a negative? No, we have dissociation. Constant will be in negative concentration must be played with hydro Nia Mind concentration divided by the actor concentration It's a no for the acid. The association off Etch a can be proceeds as ich 30. We have Etch A and we have concentration off h 30 positive equals villain to concentration off in negative. So k a. They can also be written us etch three or positive hold square because the concentration is equivalent divided by reacting concentration at a no, we can determine the hydro Nia mind concentration by rearranging the equation. Else get a much deployed with a J concentration the fraction dissociation so fraction dissociation can be represented US concentration off reacting, dissociated derided by the total reacting equal toe I don't mind concentration divided by the act in concentration substituting the values for fraction association we have under route Okay, concentration off. React in divided by etch a concentration. This can also return us taking multiplied with the change divided by it's a what is clear for the solving. It can be represented us. It's a now for percentage dissociation. We just multiply this equation with 100 so person, the association is directly proportional to the square. Root off a and inversely proportional to the square root off initial concentration off asset.

There are several ways that you can answer this problem and one would include using mass and charge balance equations which were not introduced in this chapter. So I'm going to do what I can with what was introduced in this chapter. If we have to, Equilibrium has written here, then both of those equilibrium need to be satisfied simultaneously. And we could also state that the sum of the two equilibrium also need to be satisfied. So the sum of the two equilibrium is this equilibrium here with a K value of 4.0 times tend the negative eight. So if we have an initial concentration of H A of one Moeller then is that this equilibrium shifts. It will decrease by two X while the hydro knee, um, and the H two um h a to minus will increase by X. So then we can solve for our X value. You've seen the quadratic formula. I will get an X equal to 2.0 times 10 the negative four which will be both of these concentrations then to solve her age. A. We simply recognize that a J is one minus two times X and then the last step would be to use this K expression rearrange it to solve for a minus and a minus is going to be equal to K A multiplied by the H, A concentration divided by the hydro knee. Um, concentration and we get five times 10 the negative five Mueller.

In the ocean were provided with the weak acid, which has an ionization constant Que es equals so acid is H eight having an ionization constant K a equal 1.0 multiplied to attend a super minus eight. It also forms Etch a two negative. So it also forms Etch a toe Negative as the reaction. Let me represent it in form off Reaction's edge, eh? Plus a negative gives me etch a too negative. The equilibrium constant for this reaction that Kay turns out to be four points. You calculating the concentration off each positive, the negative, and h a two in one molar solution off a J. The ionization off h ey can be represented us. It's a yeah, what? Which gives me a negative last hydro near mine really present this change in concentration using an I C E table substituting the value from I c e table. We have equation number two. Now the initial concentration given is one Mouna. Therefore, we have equation number three and four which are equal to each other. The subsidies these two into equation number four to deter mined the value off equilibrium constant that escaped the final value off. Hey, turn out to be equation number five As shown we simplify it photo in the next, substituting the values off k A. The value off k 80 no, Toby. Concentration off Export the divine divided by the reactant concentration and subtracting the concentration off h A. Therefore, the concentration off Etch A at last can be due to mine as 1.8 multiplied with 10 days to poor negative for the concentration off a negative is 4.5 multiplied with 10 days to plan negative five on the concentration off each positive I am music will do two point toe multiplied with 10 days to put negative for.

This question asked us to take a look at the equilibrium involved with the weakest solution. So we're told the acid starts at 0.20 molar concentration and it's 3% dissociated um when it reaches equilibrium. So we're asked first to calculate the hydrogen ion concentration or hydro Nehemiah in concentration of the acid. So based on the fact that it's 3.0% associated, we can rewrite that as a fraction as 0.30 or as a decimal I should say. And as a fraction then we have 0.30 hydrogen or hydro knee um ions. Those are the my dissociated ions per 0.20 Mueller concentration solution. So that's kind of the quick and easy way to determine. Then when we do the math here, hydro any of my own concentration, it's 0.6 and we'll give that extra significant figure there. So we have to sig figs, Mueller hydro knee um ion concentration. So we carry out the math, we find that the hydro any of my own concentration is again a 0.6 moller. When we put that into play for the definition of ph which is minus the log of that hydro me um concentration. We get the ph value for that solution would be to 0.22 Now we're asked to also determine the P. O. H. And hydroxide ion concentration of the solution. So we need to use a couple of relationships for Equus solutions to help us with that ph plus P O. H equals 14. So if we plug in the value for the ph here at 2.22 we can solve for P. O. H. And determined that that is 11.78 The final step is then asking us to determine the hydroxide ion concentration of the solution. Well, again, by definition P. O H is minus the log of hydroxide iron concentration and we know that the pio ages 11.78 So we'll carry the negative over. And if we solve for the log of the hydroxide iron concentration as minus 11.78 and take the inverse log of this number, we should be able to find out the hydroxide and concentration. So when we take that inverse log, we find out that the hydroxide ion concentration in the solution is 1.7 times 10 of the negative 12 molar. So now we have solved for all the things we need to know in part a part B of the question that asks us to determine the K. A. Value for this weak acid. So we're going to take a look at setting up the K expression that's going to be, we refer to this as the hydro any of my own concentration and the conjugate base of this acid. We're just going to refer to it as a minus. Okay. And the under associated form of this week acid we're going to refer to as H. A. What we know is that for the equilibrium or for the equilibrium expression K. A. Value that both the hydrogen my own concentration and the conjugate base are going to be the same value. So we know those two values are going to be point 0060 We had that number from above. Okay. We also know the starting concentration of our acid was 0.20 And from that we need to subtract off the 0.60 that dissociated into ions. If we solve this problem will have the K. A. Value for the assets carrying out the remaining steps and kind of simplifying the equation and solving it to two significant figures that were limited to the K value. Then, for this acid is 1.9 times 10 to the negative four.


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