Rishi's Chemistry Thread

[lzxnl]

When methane reacts with chlorine in the presence of uv light,
a) an addition reaction occurs to form CH₃Cl
b) a substitution reaction occurs to form CH₃Cl
c) the products are CH₃Cl and HCl only
d) the products include CH₃Cl, CH₂Cl₂, CHCl₃, CCl₄ and HCl

Troll question. b can occur, but d is more accurate because of all the possible products. Note that HCl IS a product of the substitution, no matter if it's the first, second or third stage.

[Rishi97]

From the following statement, is it possible to tell the direction in which the rate of reaction is greater?
---> The concentration fraction is greater than the value of the equilibrium constant

[jgoudie]

Yes.  The concentration factor is the Q value (ie. [products]/[reactants]) at any time during an equation.  The equlibrium constant is the K value (ie. [products]/[reactants]) at equilibrium.

Thinking about this.  If the Q value is larger than the K value that means there is more products than there would be at equilibrium. These means that the reaction would proceed in the reverse direction.

From the following statement, is it possible to tell the direction in which the rate of reaction is greater?
---> The concentration fraction is greater than the value of the equilibrium constant

[Rishi97]

An aqueous solution was prepared by dissolving 0.0200g of NaOH in 500.0mL of deionized water. The pH of this solution would be closest to:
a) 11.0
b) 3.0
c) 3.3
d) 10.7

Thank you heaps

[alchemy]

An aqueous solution was prepared by dissolving 0.0200g of NaOH in 500.0mL of deionized water. The pH of this solution would be closest to:
a) 11.0
b) 3.0
c) 3.3
d) 10.7

Thank you heaps

n=0.02/(23+16+1)=0.0005
c(OH-)=[OH-]=n/V=0.0005/0.5=0.001M
[H+]=(10^-14)/0.001=10^-11
pH=-log₁₀[H+]=-log₁₀(10^-11)=11.
So the answer is A.

[Yacoubb]

An aqueous solution was prepared by dissolving 0.0200g of NaOH in 500.0mL of deionized water. The pH of this solution would be closest to:
a) 11.0
b) 3.0
c) 3.3
d) 10.7

Thank you heaps

Before you do any calculations, just keep in mind that sodium hydroxide is a strong base, eliminating B and C. Then, calculations as alchemy has shown.

[Rishi97]

The addition of some solid sodium propanoate to a 0.10M aqueous solution of propanoic acid resulted in the equilibrium concentration of the propanoate ion being 5.2 x10^-3. The pH of this solution would be:

a) 3.6
b) 2.9
c) 2.3
d) 6.2

[alchemy]

The addition of some solid sodium propanoate to a 0.10M aqueous solution of propanoic acid resulted in the equilibrium concentration of the propanoate ion being 5.2 x10^-3. The pH of this solution would be:

a) 3.6
b) 2.9
c) 2.3
d) 6.2

Use the Henderson-Hasselbalch equation.
pH=pKa (of propanoic acid) + log(propanoate concentration/propanoic acid concentration)
=-log(1.3*10^-5) + log((5.2*10^-3)/(0.1))=3.6.
So the answer is A.

[Blondie21]

Use the Henderson-Hasselbalch equation.
pH=pKa (of propanoic acid) + log(propanoate concentration/propanoic acid concentration)
=-log(1.3*10^-5) + log((5.2*10^-3)/(0.1))=3.6.
So the answer is A.

Hey alchemy, which textbook do you use? I have never heard of that equation before.

-

I solved this differently and used acidity constants (Rishi it's on page 289 of Heineman)

CH₃CH₂COOH <-> CH₃CH₂COO- + H+

Ka = [CH₃CH₂COO-] [H+ or H3O+.. same thing] / [CH₃CH₂COOH]

Sub in the info we now. NB: the acidity constant for propanoic acid is in our data books

1.3 x 10^-5 = [5.2x10^-3] x [H+] / 0.10

H+ = 0.00025M

pH = -log(0.00025)
pH = 3.6

(My class did this last week )

[Rishi97]

Hey alchemy, which textbook do you use? I have never heard of that equation before.

-

I solved this differently and used acidity constants (Rishi it's on page 289 of Heineman)

CH₃CH₂COOH <-> CH₃CH₂COO- + H+

Ka = [CH₃CH₂COO-] [H+ or H3O+.. same thing] / [CH₃CH₂COOH]

Sub in the info we now. NB: the acidity constant for propanoic acid is in our data books

1.3 x 10^-5 = [5.2x10^-3] x [H+] / 0.10

H+ = 0.00025M

pH = -log(0.00025)
pH = 3.6

(My class did this last week )

Haha thanks Blondie 
Yeah this formula sounds more familiar hehe

[Rishi97]

Use the Henderson-Hasselbalch equation.
pH=pKa (of propanoic acid) + log(propanoate concentration/propanoic acid concentration)
=-log(1.3*10^-5) + log((5.2*10^-3)/(0.1))=3.6.
So the answer is A.

Never seen this before..hmm.. something new for me:) yay
Thanks Alchemy

[Blondie21]

Never seen this before..hmm.. something new for me:) yay

Me too hehe yay!

Since it's not in the Chemistry study design, are we still allowed to use this equation for questions?

[lzxnl]

Use the Henderson-Hasselbalch equation.
pH=pKa (of propanoic acid) + log(propanoate concentration/propanoic acid concentration)
=-log(1.3*10^-5) + log((5.2*10^-3)/(0.1))=3.6.
So the answer is A.

lol random equation from uni chem much?

Me too hehe yay!

Since it's not in the Chemistry study design, are we still allowed to use this equation for questions?

It's not part of VCE, don't worry.
Besides it's actually nothing special.

The Henderson-Hasselbalch equation is pH = pKa + log([base]/[acid])
where pKa = -log(Ka)
All logs are base 10

It's very easy to prove if you assume [base] and [acid] at equilibrium are their initial values, which is the requirement for using this equation.

Ka = [H+][base]/[acid]
log(Ka) = log([H+]) + log([base]/[acid])
-log([H+]) = -log(Ka) + log([base]/[acid])
pH = pKa + log([base]/[acid])

So yeah...nothing special, just a shorthand

[Rishi97]

A student titrates sulfuric acid against a 0.200 M sodium hydroxide solution. 20.00 ml aliquots of sodium hydroxide are used
How do I know if the acid or the base is in the burette?

[alchemy]

Hey alchemy, which textbook do you use? I have never heard of that equation

Uhh, I've been reading way too much of Wikipedia.