Past Paper Solutions Request & Mole Calculation (1 Viewer)

[dno]

Hi there.

I was just wondering whether anyone has solutions to the 2006 and 2007 past hsc papers for physics and/or chemistry or if anyone could upload them. If not could someone help me with a certain question.

Its on the 2007 Chemistry HSC paper (question 21 c), im usually good at calculations but cant seem to solve this, my tutor cant solve it either.

Well anyway heres the question,




What volume of 0.005mL^-1 KOH is required to neutralise 15mL of the diluted solution of H₂SO₄?<o></o>

Thanx
<o></o>

 

[minijumbuk]

21.
a) Assuming that both the 1st and 2nd ionisations of H₂SO₄ are strong, then [H⁺]= 0.005 x 2 =0.01M
pH= -log₁₀10^-2
= 2
Therefore colour of red cabbage solution in the 0.005M sulfuric acid will be red

b)Moles of sulfuric acid = 0.005x0.01 = 5x10^-5 mol
If we dilute to 100 mL, then the diluted solution will have concentration of:
(5x10^-5 ) / 0.1
=5x10^-4 M
So [H⁺] = 10^-3 M
pH = -log₁₀10^-3
= 3

Therefore violet colour

c)Diluted sulfuric acid, from part b), has [H⁺] of 10^-3 M
Moles of H⁺ will therefore have 0.015 x 10^-3 mol = 1.5x10^-5 mol
Therefore moles of OH^-= 1.5x10^-5 mol
Volume of KOH = 1.5x10^-5 / 0.005
= 3x10^-3 L
= 3mL


By the way, you should get a new tutor.

 

[dno]

thanx for ur advice.
I was in doubt, but i wasnt sure, that we should assume that its the same concentration of the one from part b. but i can understand that.
Thanx, i know how to do it now, but i think u made a mistake.
If u write out the chemical reaction equation, you'll find the moles of KOH is double the moles of Sulphuric Acid (ie. 2:1 ratio), therefore u have to double the amount of moles u have used in ur calculations.
When i done it i got 6mL
Please give me ur comments, i hope im right :wave:

 

[xiao1985]

It was already taken into account I believe

(concentration of H2SO4) =5x10-4 M
So [H+] = 10-3 M

 

[minijumbuk]

What xiao said