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#1 ginneswatson

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Posted 30 May 2009 - 12:28 PM

Just introducing myself. I'm a full time teacher of Chemistry who stumbled upon this site a few years ago. Greatly admired the way you all help each other but couldn't resist correcting some errors. Fortunately everyone seemed more than happy so I've returned most years at about this time to offer help over the last few days of your preparation.

This Advanced Higher Forum was really busy so it is disappointing that it seems to have 'died a death'. However, if there are people looking in - don't be shy!

I'll try and look back regularly and give what help I can.

Cheers and good luck! blink.gif

#2 123Chemistry

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Posted 01 June 2009 - 09:48 PM

QUOTE (ginneswatson @ May 30 2009, 01:28 PM) <{POST_SNAPBACK}>
Just introducing myself. I'm a full time teacher of Chemistry who stumbled upon this site a few years ago. Greatly admired the way you all help each other but couldn't resist correcting some errors. Fortunately everyone seemed more than happy so I've returned most years at about this time to offer help over the last few days of your preparation.

This Advanced Higher Forum was really busy so it is disappointing that it seems to have 'died a death'. However, if there are people looking in - don't be shy!

I'll try and look back regularly and give what help I can.

Cheers and good luck! blink.gif


Hi there. Please could you help me with multiple choice questions 18 and 22 from the 2005 paper?
I am spending hours trying to figure them out but can't!!
18) For any acid base indicator, the colour change occurs around pH=pKin. The equilibrium constant Kin, of an indicator used in an acid-base titration where the pH at the end point was 5.2, is
A)6.3^-6 (the correct answer)
B)0.72
C)5.2
D)1.6^-5

22) Tin can exist in two different forms, white tin and grey tin. For the change white tin to grey tin, standard enthalpy change is 2.5kJ mol -1 and standard entropy chenge is -6.7JK-1mol-1,
and hence standard free energy change (G) at 298 K in Kjmol-1 will be,
A)-0.5 (correct answer)
B)-4.2
C)4.5
D)9.2
Any help would be much appreciated!!


#3 ginneswatson

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Posted 02 June 2009 - 06:11 AM

Hi there. Please could you help me with multiple choice questions 18 and 22 from the 2005 paper?
I am spending hours trying to figure them out but can't!!
18) For any acid base indicator, the colour change occurs around pH=pKin. The equilibrium constant Kin, of an indicator used in an acid-base titration where the pH at the end point was 5.2, is
A)6.3^-6 (the correct answer)
B)0.72
C)5.2
D)1.6^-5

22) Tin can exist in two different forms, white tin and grey tin. For the change white tin to grey tin, standard enthalpy change is 2.5kJ mol -1 and standard entropy chenge is -6.7JK-1mol-1,
and hence standard free energy change (G) at 298 K in Kjmol-1 will be,
A)-0.5 (correct answer)
B)-4.2
C)4.5
D)9.2
Any help would be much appreciated!!
[/quote]

Q18 Should just be a case of converting pKIn to KIn.

pKIn = - log ( KIn) so KIn = 10 -pKIn

Use your calculator to put in 10 -5.2 and I presume it will tidy it up as 6.3 x 10-6

Q22 Most common mistake is forgetting that Enthalpy is in kiloJoules while Entropies are usually just Joules

G = H -TS so G = (-2,500 - (298 x -6.7)) = -2500 +1996.6 = -503.4 J = -0.5034 kJ

(To get the right answer, I've had to assume that Enthalpy is exothermic (-ve) and that you missed out the minus sign in the original question. Maybe that is where you went wrong originally?)


#4 neep

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Posted 02 June 2009 - 08:52 AM

hi i was wondering if you could help me with 2008 question 15
500cm^3 of 0.022moll^-1 hydrochloric acid was mixed with 500cm^3 of 0.02moll^-1 sodium hydroxide solution, the Ph will be
A 2
B 3 correct answer
C 4
D 5
Thanks

#5 ginneswatson

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Posted 02 June 2009 - 10:42 AM

QUOTE (neep @ Jun 2 2009, 09:52 AM) <{POST_SNAPBACK}>
hi i was wondering if you could help me with 2008 question 15
500cm^3 of 0.022moll^-1 hydrochloric acid was mixed with 500cm^3 of 0.02moll^-1 sodium hydroxide solution, the Ph will be
A 2
B 3 correct answer
C 4
D 5
Thanks


Need to work out moles of H+ and moles of OH- and then work out which one will be left over (in excess)

H+, n = C x V = 0.022 x 0.5 = 0.011 mol

OH-, n = C x V = 0.020 x 0.5 = 0.010 mol

H+ in excess by 0.011 - 0.010 = 0.001 mol

Total volume = 500 cm3 + 500 cm3 = 1 litre

Concentration of H+ = n / V = 0.001 / 1 = 0.001 mol l-1 = 10-3 mol l-1

so pH = 3

There are 'shortcuts' but that gives you the detailed path.

ps. Could I suggest that you start a new thread for each question so that other users realise that this not just about "Help Available". Cheers tongue.gif

#6 neep

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Posted 02 June 2009 - 12:31 PM

Thank you.





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