 # Chapter 5: Solution Chemistry

### Re: Chapter 5: Solution Chemistry

zimengy17106 wrote:
In Question 19, "A solution of 50 gram of KCl at 85 degree was slowly cooled to 60 degree without precipitate". But the volume/weight of solution is unknown. The online solutions choose D, but it is based on the assumption that it is 50g KCl in 100g water. However, if it is 50g KCl in 119g water (which equals 42g KCl in 100g water), then at 85 degree it is unsaturated and at 60 degree it is just saturated without any precipitate. In this way, Choice A is also correct?

That is not a correct interpretation of the information provided.

The definition of supersaturated is given in Table 1: more solute than a saturated solution.

50 g at 60 degrees, no precipitate, means that 50 g is in solution which is above the saturation curve at 60 degrees, by definition, supersaturated.
goldstanda3269

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### Re: Chapter 5: Solution Chemistry

zimengy17106 wrote:In Question 29, I thought choice C is also correct? Because when A and B contribute equally, which is at the joint point of the 2 green lines, on the x axis it looks like XA=0.3 XB=0.7, so component B is in greater concentration than A. Why is C wrong?

You are correct! Answer choices C and D present correct statements so give yourself an extra mark!

This will be updated on Tuesday. Keep up the good progress with your studies!
goldstanda3269

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### Re: Chapter 5: Solution Chemistry

This is regarding GEN CHM Chapter 5 Worked Solution for Question 24:

Why is it that in this question we are required to calculate constant k? I thought constant k for any equation is some fixed number with appropriate units that is usually provided and then you calculate things based on info provided including constant k?
I am confused of why the equation was re-arranged to k = C/P to calculate this constant? Since the numbers plugged being C or P can change it seems that the constant can change making it not so constant after all?

Regards,
Ania
horbowiec.7527

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### Re: Chapter 5: Solution Chemistry

This is regarding GEN CHM Chapter 5 Question 29 (CHM-137) & Worked Solution:

I understand why answer D is considered correct, it's obvious, and that is what I chose here.

However could you please explain why answer B from the book: "The point at which both components contribute equally to the total vapor pressure, component B is in the greater concentration" is considered incorrect?

Worked solution state: "Answer choice C is incorrect, since 2 components contribute equally when 2 green lines cross, and at that point, mole fraction leans to component B side (thus B dominates at that point)."

So I am confused here, at the point that 2 green lines cross both components contribute equally and at that point component B is in greater concentration as per the position of the crossed point and info at right bottom side of graph Xb =1, Xa = 0.

Am I missing something here?

Regards,
Ania
horbowiec.7527

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### Re: Chapter 5: Solution Chemistry

horbowiec.7527 wrote:This is regarding GEN CHM Chapter 5 Worked Solution for Question 24:

Why is it that in this question we are required to calculate constant k? I thought constant k for any equation is some fixed number with appropriate units that is usually provided and then you calculate things based on info provided including constant k?
I am confused of why the equation was re-arranged to k = C/P to calculate this constant? Since the numbers plugged being C or P can change it seems that the constant can change making it not so constant after all?

This is an important question because it also applies to GAMSAT Physics and GAMSAT Biology, as well as other areas in Gen Chem (later you can see Chapter 9 questions 13-15, again with the calculation of a constant and its units).

First, in Chemistry, a constant is usually specified for a particular pressure and/or temperature (e.g. STP). In Question 24, the pressure is changing (but we have Henry's Law to deal with that) but the temperature is constant at 20 oC. So k will be the same because the temperature is not changing.

You are right that C or P can change, but, since k is constant, the ratio of C/P does not change (as long as the temperature is constant). And that means that once you have k, they can give you any C or any P and you will easily be able to calculate the missing variable.
goldstanda3269

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### Re: Chapter 5: Solution Chemistry

horbowiec.7527 wrote:This is regarding GEN CHM Chapter 5 Question 29 (CHM-137) & Worked Solution:

I understand why answer D is considered correct, it's obvious, and that is what I chose here.

However could you please explain why answer B from the book: "The point at which both components contribute equally to the total vapor pressure, component B is in the greater concentration" is considered incorrect?

Worked solution state: "Answer choice C is incorrect, since 2 components contribute equally when 2 green lines cross, and at that point, mole fraction leans to component B side (thus B dominates at that point)."

So I am confused here, at the point that 2 green lines cross both components contribute equally and at that point component B is in greater concentration as per the position of the crossed point and info at right bottom side of graph Xb =1, Xa = 0.

Am I missing something here?

The worked solution is correct; however, the January 2021 print version Masters Series book had a typo for answer choice C, component B, which was corrected to Component A in the eBook and in the June, 2021, and subsequent print editions. If you have the January 2021 print version and chose C as the correct answer, give yourself a bonus point! And even if you missed it at that time, give yourself a bonus point for noticing it later!
goldstanda3269

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### Re: Chapter 5: Solution Chemistry

Hi there,

In relation to Question 14, why is 60 grams included within the worked solutions?

Thank you so much,

Kind regards and best wishes,

Em Stevens
emstevens46670

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### Re: Chapter 5: Solution Chemistry

emstevens46670 wrote:
In relation to Question 14, why is 60 grams included within the worked solutions?

Notice that Figure 1 presents data "per 100 g", per 100 grams, of water but we are given a "compound in 50 ml of water." We must start by ensuring that the units are identical.

30 g of an unknown compound in 50 ml is the same as 60 g of an unknown compound in 100 ml (simple proportion; GM 1.6); given the density of water is 1 g/ml (many students commit this to memory but it is also in the "Note that" section above the graph), we are looking for 60 g of an unknown compound in 100 g of water. Now we can use Figure 1.
goldstanda3269

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### Re: Chapter 5: Solution Chemistry

Hi,

Regarding page 127, i noticed that when calculating the Ksp, the concentration must be raised to the power of the ions coefficient , for example

Ag+ becomes Ksp = [x]
whereas 2Ag+ = [2x]^2 just wondering..

1) why we have to account for the 2, twice once with the 2 in the 2x and then again by squaring it?
2) is this only when solving for Ksp that we have to account for the coefficient twice?

Any info would be much appreciated. Thank you!
skogias6733

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### Re: Chapter 5: Solution Chemistry

skogias6733 wrote:Hi,

Regarding page 127, i noticed that when calculating the Ksp, the concentration must be raised to the power of the ions coefficient , for example

Ag+ becomes Ksp = [x]
whereas 2Ag+ = [2x]^2 just wondering..

1) why we have to account for the 2, twice once with the 2 in the 2x and then again by squaring it?

(1) When you are doing Keq or other equilibrium constants, the square brackets [] represents concentration in mol/L. You would never "account for the 2, twice".

However, it is different for Ksp and the reason is shown in the step-by-step calculation in the book.

This is more than you need, but if you want a different way to look at the calculations: https://studylib.net/doc/8131747/proble ... g-with-ksp

skogias6733 wrote:2) is this only when solving for Ksp that we have to account for the coefficient twice?

(2) You do not need to memorize this information because, if needed during the exam, the information will be provided and often provided with an example.

Nonetheless, this type of calculation for GAMSAT is unique to Ksp. However, it is not as simple as saying Ksp "account for the 2, twice". Why? It depends on what you are calculating.

Either:
1) You are given the actual concentration, in which case, that is the ONLY number that you use. You would not multiply by 2 but you would raise to the power of 2. Notice that in the equation that you put in your post, that you inadvertently replaced the s for solubility (from the book) with x which often represents the actual concentration. {Side note: Of course, it never matters what letter you use to represent an idea as long as you know what that letter represents.}
2) You are trying to determine the solubility of an ion, then you need to "account for the 2, twice" (or the 3, or whatever the stochiometric coefficient is).

PS: You are doing great! Keep up the good work!
goldstanda3269

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### Re: Chapter 5: Solution Chemistry

Hi,

Thanks for that, that document was useful.

But what if you have the initial concentrations of a reaction and want to find the equilibrium concentrations using the ICE table and one of the reactants has double the stoichiometry? For example 2HI > H2 + I2

so then putting in the equilibrium values from the ICE table values into the Keq, wouldn't you also put 2x^2 here? if the stoichiometry is double?
e.g. 2HI > H2 + I2
I 0.8 0 0
C - 2x +x +x
E (0.800 - 2x) x x

2HI > H2 + I2 = [prod]/[reactants] = x^2/ [0.800 - 2x]^2

Thanks for the help I appreciate it!
skogias6733

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### Re: Chapter 5: Solution Chemistry

skogias6733 wrote:
But what if you have the initial concentrations of a reaction and want to find the equilibrium concentrations using the ICE table and one of the reactants has double the stoichiometry? For example 2HI > H2 + I2

so then putting in the equilibrium values from the ICE table values into the Keq, wouldn't you also put 2x^2 here? if the stoichiometry is double?

Definitely not! This is the difference between calculating the actual concentration at a given moment (which is the situation that you are describing) versus calculating the solubility at a given temperature and pressure (which is affected by 2x in your example).

To be clear, we are talking about a problem where ACER would only go this far if a clear example is provided to use as a template to solve the problem. There is nothing to memorize here for GAMSAT purposes. Understanding how to perform the calculations (the maths) is critical, but understanding how to set up the calculation without a template is not going to happen for this exam. I'm trying to be sure you are not trying to focus on "should I multiply or not by the coefficient" when that will never be the concern on the real GAMSAT. Of course, the end-of-chapter practice questions and practice exams will ensure that you have a clear insight as to what is required.
goldstanda3269

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### Re: Chapter 5: Solution Chemistry

Ok great thank you!!
skogias6733

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Joined: Sat Feb 06, 2021 11:17 am

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