A list of multiple choice questions :::: (1 Viewer)

[norelle]

These are the multiple choice questions that I'm unsure with..
Please help
and there is a calculation question that I dont understand as well (q.10)
Any help would be appreciated ~~ THANKS!!!


1) Acid Conc (mol/L) pH
P 0.01 2
Q 0.05 1
R 0.1 1
S 0.1 2

Which acid can donate more than one proton?
ANS: Q
why?
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2) Given 0.1M aqueous solutions of each of the following listen compounds, which would be the most basic?
A) NH3
B) NaOH
C) Ca(OH)2
D) NaCH3COO

Some say B, some say C, which one is correct?

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3) Which of the following species could best be described as amphiprotic?
A) HNO3
B) NH2-
C) H2PO4-
D) CH3COO-

ANS: C
Why?
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4) 300mL of 0.45M HCl reacted with 15g of sodium sulfite to form NaCl, SO2, H2O.
What volume of SO2 was produeced at 25*C amd 100kPA?
A) 1.7L
B) 2.7L
C) 3.0L
D) 3.3L
ANS: C
But I got A ....
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5) Which correctly identifies a Bronsted-Lowry acid-base pair?

Acid Base
A) H2F2 HF
B) H2O (OH)-
C) (HCO3)- H2CO3
D) CH3COOH CH3OH

ANS: B
why cant be C?
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6) A student reacts equal amounts of 5M NaOH solution, and 5M HCl at room temp.
A) temp before reaction = temp after reaction
B) temp before reaction < temp after reaction
C) temp before reaction > temp after reaction
D) temp before reaction > temp after reaction(much lower)

ANS: B why?
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7) Which of the following pairs would form a buffer solution?
A) H2SO4/(HSO4)-
B) OH-/H2O
C) HCOOH/HCOO-
D) HNO3/NO3-

ANS:C
buffer solution can either be weak acid/weak base and its salt.
so I can cross out A and D, the answer is C.. but why cant be B?
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8) The indicator used for titration required the pH to be maintained at about 10.
Which of the following mixtures is suitable as a buffer solution?
A) Ammonium chloride, HCl
B) NAOH, sodium ethanoate
C) Sodium ethanoate, CH3COOH
D) Ammonia and ammonium chloride

ANS: D
I know C & D are buffer as they are weak acids, but how do I know which one is more acidic/basic?
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9) H2CO3 & NaHCO3 AND NaHCO3 & Na2CO3 (this one more basic)
Which buffer is more acidic?
How do you know?
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10)
A student wished to determine the percentage of CaCO3 present in a shell found at the beach.
The clen dry shell, which weighed 1.306g, was placed in a small beaker and 10mL of 5M HCl was added.
When the shell had completely dissolved, the resulting solution was transferred to a volumetric flask
and the volume made up to 25mL with distilled water.
A 10 mL sample from this solution required 11.2mL of 1M sodium hydroxide for complete neutralisation.

CaCO3 + 2HCl -> CaCl2 + CO2 + H2O

a) Calculate the number of moles of NaOH present in the 11.2 mL of 1M NaOH solution
this one I got 0.0112 moles

b) How many moles of acid remained in the beaker after the reaction with the shell (before the dilution was made)?
ANS: 2.8 x 10^-2 moles

c) How many moles of acid reacted with the shell?
ANS: 2.2 x 10^-2 moles

d) Calculate the mass and the percentage mass of CaCO3 was present in the shell
1.10g / 84.2%

 

[Pwnage101]

1)pH=-log[H+]

now if an acid can donate 2 protons [H+] will be TWICE [Acid]

thus we are looking for the one where the pH and acid concentration match:

pH=-log(2[Acid]) which is Q, since -log(2x0.050= -log(0.1) = 1 which = pH given

2)Id say C since it produces twice the amount of [OH-] and is a strong base.

3) amphiprotic means able to both donate and accept a proton

here are the optionsa nd why they are correct/incorrect:
A) HNO3 -INCORRECT - can donate a proton, but will not accept, [H2NO3+ wont be produced]
B) NH2- - INCORREC- will accept a proton, but will not donate [NH(2-) will not form]
C) H2PO4- - CORRECT = donates a proton to form HPO4(2-) and accepts to form H3PO4
D) CH3COO- - INCORRECT - willa ccept proton, but will not donate any - the three hydrogens in the structure arew not replacable.

4)300mL of 0.45M HCl reacted with 15g of sodium sulfite to form NaCl, SO2, H2O.
What volume of SO2 was produeced at 25*C amd 100kPA?

limiting reacgent - watch out for these

equ is 2HCl (aq) + Na2SO3 (s) ------> 2NaCl (aq) + SO2 (g) + H2O (l)


for HCl we have moles = n = cV = 0.3 x 0.45 = 0.135 moles

for Na2SO3 we have moles = n = m/M = 15/126.04 = 0.119009838 moles

we need HCl: Na2SO3 in a 2:1 ratio, thus the limiting reacgent is the HCl

thus we use this value for n (0.135 moles)

thus moles of SO2 produced IS 0.135/2 =0.0675 moles (since HCl: SO2 is a 2:! ratio)

n = v/V where V = molar volume, rearrange and v = n x V = 0.0675 x 24.79 = 1.673325 L

so i would also say A, i think the answer is wrong, they used Na2SO3 as the limiting reagent, which it is not. so YOU are correct.

5) conjugate acid/base pairs differ by a H+ only

here are the options and why theya re correct/incorrect
Acid Base
A) H2F2 HF - INCORRECT -doesnt differ by one H+
B) H2O (OH)- - CORRECT - differs by 1 H+
C) (HCO3)- H2CO3 - INCORRECT - the heading states "acid/base" so teh acid must coime first, in this case the base comes first under teh acid column and so is incorrect - H2CO3 is the acid. If the columns were switched it would be correct
D) CH3COOH CH3OH - INCORRECT - does not differ by one H+

6) this is a neutralisation reaction between a strong abse and stronga cid, therefore HIGHLY exothermic, which by definition means heat is produced i.e. temp before < temp after.

7) Cant be B) beacuse i OH- is a STRONG base, we prefer to have something weaker as our conjugate. Also H2O is amphiprotic and could form H3O+.

8) comes with practice. Now u know

9)Well H2CO3 is acidic, NaHCO3 is slightly basic, and Na2CO3 is basic. Thus its pretty obvious which 2 choose for the most acidic buffer.

10) ill do a little later

 

[norelle]

Thanks so much.
Thankyou for being so helpful

 

[rufous]

Since NaOH and HCl react on a 1:1 ratio (write the equation to verify), then the number of moles of unreacted HCl in the 10 mL sample is the same as answer a). BUT, since there was 25 mL of diluted solution altogether we must multiply the 0.0112 by 2.5 to find the total moles of excess HCl.

The moles of HCl reacting can be found by subtracting the excess HCl from the number of moles of HCl added in the first place. ie .050 - .028 = .022 mol.

Now, from the supplied equation, the number of moles CaCO3 reacting is HALF this number of moles ie. .011 mol. and since the approx. formula mass of CaCO3 is 100 amu , we can find its mass by multiplying the two as n = m/M giving 1.100 g

Since the original mass of the shell was 1.306, the % CaCO3 present by mass is : 1.100/1.306 x 100% = 84.23%

This question assumes that CaCO3 is the only thing present in the shell which reacts with the acid!!!!!