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Chapter 6: Acids and Bases

A question about the book page CHM-89

Postby li.zhang96890 » Fri Oct 16, 2015 2:48 pm

On the book page CHM-89, it says ' The midpoint of the titration is the equivalence point of the titration curve' in the bracket. But I think the equivalence point is the end point basically, where no NH3 exists, and at the mid point of the titration, as the book says, [NH4+] = [NH3]. so I think that sentence is wrong. What do you think?
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Re: Chapter 6: Acids and Bases

Postby trhmd20126928 » Sat Oct 24, 2015 8:05 am

"On the book page CHM-89, it says ' The midpoint of the titration is the equivalence point of the titration curve' in the bracket. But I think the equivalence point is the end point basically, where no NH3 exists, and at the mid point of the titration, as the book says, [NH4+] = [NH3]. so I think that sentence is wrong. What do you think?"

The equivalence point refers to the point at which one hydrogen is either removed (acid titrated with base) or added (base titrated with acid). In this example, the weak base, NH3, is being titrated with the strong acid, HCL. So, the equivalence point occurs when NH3 becomes NH4. For any titration curve, the midpoint is also the equivalence point, so the statement is correct.

However, the equivalence point and the end point are not the same. The end point refers to the point at which an indicator changes color. Now, the indicator will change color around the same time that a hydrogen is added or removed. So, while the equivalence point and end point are not the same, they do occur around the same time.
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Re: Chapter 6: Acids and Bases

Postby li.zhang96890 » Tue Nov 03, 2015 2:59 pm

Sorry, the mid point is where [NH4+] = [NH3], i.e. it is where half the analyte has been titrated. But the equivalence point is the point at which chemically equivalent quantities of acid and base have been mixed. In other words, the moles of acid are equivalent to the moles of base.(https://en.wikipedia.org/wiki/Equivalence_point)
So, at the end point, all the NH3 has been titrated. Hence, the mid point and equivalence point are totally different concepts.

But I agree to you that equivalence point and the end point are not the same though they are quite close.

Look forward to your response.
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Re: Chapter 6: Acids and Bases

Postby goldstanda3269 » Sat Nov 14, 2015 7:04 am

ilbe wrote:As lewis structure of HClO3,
1. Cl is not following octect rule. Is this exceptional case?
2. Any other molecules that are not following octect rule for GAMSAT?



1. Yes, this is an exceptional case. GAMSAT does not expect this as presumed knowledge. Just a side note: There are 3 resonance structures for HClO3, 1 follows the octet rule and 2 do not: http://chem-net.blogspot.ca/2012/01/lew ... le_13.html


2. There are hundreds of exceptions (http://chemwiki.ucdavis.edu/Theoretical ... Octet_Rule). They do not represent presumed knowledge for the GAMSAT.
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Re: Chapter 6: Acids and Bases

Postby goldstanda3269 » Sun Dec 13, 2015 5:06 am

ilbe wrote:Queries with regards to H-H equation.

AS per 6.8 H-H equation is
pH=pKa + log( ['salt']/[acid])

1.The 'salt' is same as conjugate base?
2. If pOH = pKb + (['salt]/[base]), the salt will be conjugate acid?
3. Then can I make conclusion that the 'salt' in H-H equation means 'conjugate acid or base'.


1. No. From your review of Solution Chemistry (Chap 5) and Acids/Bases (Chap 6), a conjugate base often has a negative charge (i.e. bicarbonate, HCO3-; Cl-) whereas a salt is normally neutral (i.e. sodium bicarbonate, NaHCO3; NaCl). The 2 terms are NOT interchangeable. However, the concentrations of the salt and conjugate base may be the same and that’s why the magnitude of their concentrations may be interchangeable in the H-H equation (only because the values are the same).
2. The salt is the salt. Whether or not the salt has the same concentration of the conjugate acid depends on the problem.
3. No.
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Re: Chapter 6: Acids and Bases

Postby goldstanda3269 » Mon Dec 14, 2015 10:27 am

ilbe wrote:The reason why I asked was ACER Question 104-107 in Pruple booklet.
The given passage in that question stated, '[A-] is the equilibrium concentration of the conjugate base of the weak acid.' This make me confused because it is different from GS textbook.


Yes, [A-] is definitely the equilibrium concentration of the conjugate base of a weak acid (HA).

Please consider re-reading GS book CHM 6.3, 6.6, 6.7 and 6.8. Dr. Ferdinand made videos presenting all the solutions of all of ACER's practice materials and THEN added 400 pages to make the current edition (2015-2016) of the GS book. There is no difference in the science in the GS book and ACER.


ilbe wrote:So 'Salt' is more correct expression for H-H equation?


This is not what I've been saying. You must follow the directions in the passage that they provide to you. They may say salt, they can say anything else, they can even invent something new or use a definition that they don't expect that you have seen before. This is the nature of the GAMSAT. It is primarily a reasoning test with basic assumed knowledge (you MUST understand a salt, conjugate acid/base, acid/base strengths, solubility, and the rest of the information described in the first page of Chapters 5 and 6 in order to be able to apply reasoning to solve these problems).
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Re: Chapter 6: Acids and Bases

Postby ElleRT » Sun Jun 19, 2016 6:32 pm

I'm just a little confused with this question: pka acetic acid is 5. What is the pH at which the concentration of acetic acid will be 10^3 of its conjugate base, acetate?

Why do we use the hassel-bach equation. And why is [A-]/[AH] = 10^-3?

Thanks
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Re: Chapter 6: Acids and Bases

Postby ElleRT » Sun Jun 19, 2016 6:44 pm

Also for Q.11 Suppose that 30mL of 0.01M HClO4 is added to 10mL of 0.03M NaOH. What is the pH of the resulting solution?

I don't really understand the explanation, have they just worked out the mole of each compound? I'm just not sure how they got to that sum. Thanks
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Re: Chapter 6: Acids and Bases

Postby goldstanda3269 » Mon Jun 20, 2016 12:53 am

ElleRT wrote:I'm just a little confused with this question: pka acetic acid is 5. What is the pH at which the concentration of acetic acid will be 10^3 of its conjugate base, acetate?

Why do we use the hassel-bach equation. And why is [A-]/[AH] = 10^-3?

Thanks


Exactly what question are you referring to?
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Re: Chapter 6: Acids and Bases

Postby goldstanda3269 » Mon Jun 20, 2016 1:01 am

ElleRT wrote:Also for Q.11 Suppose that 30mL of 0.01M HClO4 is added to 10mL of 0.03M NaOH. What is the pH of the resulting solution?

I don't really understand the explanation, have they just worked out the mole of each compound? I'm just not sure how they got to that sum. Thanks


I think you are referring to Question 12.

So let's take a step back. The moment that you are asked about pH, all your thoughts go to:

pH = - log [H+]

so we need to find out how much H+ is floating around in solution.

HClO4 is an acid (see CHM 6.1). So the explanation shows how many moles of H+ is produced from HClO4.

NaOH is the most famous strong base in Chemistry, and the explanation shows how to calculate how many moles of OH- is produced from NaOH.

Since the number of moles of H+ is exactly equal to the number of moles of OH-, then:

H+ + OH- ---> H2O

H2O is neutral (pH 7).
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Practise Test 2 Q105

Postby ElleRT » Sat Sep 10, 2016 6:57 pm

Hi, I didn't know where else to ask but I am just going over the answers for Practice test 2 and I'm finding it quite confusing how you figure out concentration.

In no.104-107. Q.105: We are given 40.0ml of 0.25 NaH2PO4 and 25ml of 0.2M of Na2HPO4.

Why is it that we decide to figure out the mols to plug into log(conjugate base/acid). Can we not we just use the molarity?
I don't really understand why it's then divided by volume. I know the question
mol = conc x vol / L
and so what was done 0.04 L x 0.25 mol/L = 0.01 mol but if the L cancel out does this still equate to mol x l?

I only assume we can use molarity as concentration because we used it in Q34 to find the pOH of NaOH. Where we found the mol = 2mol / 0.1L = 20mol/L this was given as the concentration of pOH.

As you can see I am a little confused so any feedback would be helpful
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Re: Chapter 6: Acids and Bases

Postby trhmd20126928 » Mon Sep 19, 2016 7:09 am

[quote="ElleRT"]Hi, I didn't know where else to ask but I am just going over the answers for Practice test 2 and I'm finding it quite confusing how you figure out concentration.

In no.104-107. Q.105: We are given 40.0ml of 0.25 NaH2PO4 and 25ml of 0.2M of Na2HPO4.

Why is it that we decide to figure out the mols to plug into log(conjugate base/acid). Can we not we just use the molarity?
I don't really understand why it's then divided by volume. I know the question
mol = conc x vol / L
and so what was done 0.04 L x 0.25 mol/L = 0.01 mol but if the L cancel out does this still equate to mol x l?

I only assume we can use molarity as concentration because we used it in Q34 to find the pOH of NaOH. Where we found the mol = 2mol / 0.1L = 20mol/L this was given as the concentration of pOH.

As you can see I am a little confused so any feedback would be helpful[/quote]

Hello ElleRT,

For question 105, we must first establish common units. Since molarity (M) is in units of moles per liter (mol/L), we convert both 40.0 mL and 25.0 mL to L:

40.0 mL/1000 = 0.04 L
25.0 mL/1000 = 0.025 L

Next, we calculate the number of moles for sodium dihydrogen phosphate (NAH2PO4) and disodium hydrogen phosphate (NA2HPO4):

0.04 L (0.250 mol/L) = 0.01 mol sodium dihydrogen phosphate [weak acid]
0.025 L (0.200 mol/L) = 0.005 mol disodium hydrogen phosphate [conjugate base]

*Note that NAH2PO4 is the weak acid because it is the protonated form (extra hydrogen) and NA2HPO4 is the conjugate base because it is the deprotonated form (missing hydrogen).

Now that we have the molar concentrations, we can plug them into the Henderson-Hasselbach equation:

pH = pKa + log [0.005]/[0.01]
pH = pKa + log [.5]/[1]
pH = pKa + log (1/2)

According to log rules, log (1/X) = log X^-1 = -log X; so,

pH = pKa + log (1/2) becomes pH = pKa - log 2

The passage tells you that the pKa of H2PO4- = 7.21 and log 2 = 0.30; so,

The pH of the resulting solution = 7.21 - 0.30 = 6.91

You may also want to check out the video of this solution here: https://www.youtube.com/watch?v=JTqWAdd ... 5YtYgb714b
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6.6.1 Calculation in textbook

Postby karencowlishaw » Tue Nov 21, 2017 5:32 pm

I’m trying to work out the calculations on CHM-81 of the textbook. Where it says x^2 + (1.75x10^-5) + (-1.75x10^-7) = 0 how do you find that c=1.75x10^-7? Where do you find values for a, b and c? And from that equation could you step through the calculations please?
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Re: 6.6.1 Calculation in textbook

Postby goldstanda3269 » Tue Nov 21, 2017 11:13 pm

Section CHM 6.6.1

Of course, prior to studying Physics or General Chemistry, completing the Gold Standard GAMSAT Maths chapters is essential. No matter how much content someone studies, ACER will find a way to present unfamiliar equations in the exam room to see how you reason through it. Our chapters and practice questions in GAMSAT Maths will help.

To your question . . .

karencowlishaw wrote:I’m trying to work out the calculations on CHM-81 of the textbook. Where it says x^2 + (1.75x10^-5) + (-1.75x10^-7) = 0 how do you find that c=1.75x10^-7? Where do you find values for a, b and c? And from that equation could you step through the calculations please?


The first line of CHM 6.6.1 states what the solutions of the quadratic equation must be (note: the equation is not to be memorised since ACER is just as likely to use some other strangely named equation that you are not expected to have seen before the exam). Notice that the given equation has the number 0 on the right hand side of the equation and there are 3 sets of variables on the left: 1) the factor 'a' times x^2, 'b' times x, and positive 'c'. Now we know the standard format for a 'quadratic equation'.

... then the formula is given for the variable x (which is like solving for x). Now we just need to identify a, b, c by looking at any equation in the same format as the quadratic equation: keep in mind that 'a' must be the number multiplied by x^2 (since there is no factor in the equation provided, 'a' must be the number 1 because 1 times anything is just that anything), and b is the number multiplied by x (b = 1.75 x 10^-5), and then we must see the format PLUS c EQUALS zero, thus c = -1.75 x 10^-8.

Then you just plug those values into the solution provided for x.

All the calculations can be completed or estimated 'by hand' and are at the level expected for GAMSAT.

Please let me know if something is unclear.

Our next free webinar will be this upcoming Sunday: www.gamsat-prep.com/GAMSAT-free-online-seminar
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Re: Chapter 6: Acids and Bases

Postby Laura_2707 » Fri Feb 12, 2021 10:24 pm

Hi

Im just wondering about question 16.
Would you be able to talk me through it in a bit more detail? I don't understand the solution given!
How do you arrive at [A-]/[HA] = 10-3 ?


Thank you!
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