1,000,000 Question Thread :D

[monokekie]

Thanks!

Why does the pH change rapidly when the equivalence point is reached?

pH is calculated on a logarithmic scale, for example, a pH of 1 will have 10 times the hydronium concentration of pH 2. So it gradually takes less amount of base to neutralise the acid as the pH increases from 1 to the equivalance point. Once the pH of the solution is near the equivalence point, the change by a factor of 10 occurs really quick, thus the pH changes rapidly. This is basically why the graph is usually very steep at the equivalence point

[kenhung123]

Don't understand sorry :S

[stonecold]

Don't understand sorry :S

i am a horrible teacher

I got it

Ken, go over acid/base reactions from the text and it should be clearer.  I think the particular section is called Bronsted Lowry Theory.  But yeah, as pH is calculated on a logarithmic scale, if you can imagine a log graph, there is the part where the gradient changes very rapidly, and there are other parts where the gradient doesn't change a great deal.  This is what monokekie was talking about in relation to why the pH changes at different rates.  It depends on the concentration of hydronium ions present in the solution.

[monokekie]

Don't understand sorry :S

i am a horrible teacher

I got it

Ken, go over acid/base reactions from the text and it should be clearer.  I think the particular section is called Bronsted Lowry Theory.  But yeah, as pH is calculated on a logarithmic scale, if you can imagine a log graph, there is the part where the gradient changes very rapidly, and there are other parts where the gradient doesn't change a great deal.  This is what monokekie was talking about in relation to why the pH changes at different rates.  It depends on the concentration of hydronium ions present in the solution.

yayyy   someone understands ~

[kenhung123]

Don't understand sorry :S

i am a horrible teacher

No I am just dumb

[kenhung123]

Don't understand sorry :S

i am a horrible teacher

I got it

Ken, go over acid/base reactions from the text and it should be clearer.  I think the particular section is called Bronsted Lowry Theory.  But yeah, as pH is calculated on a logarithmic scale, if you can imagine a log graph, there is the part where the gradient changes very rapidly, and there are other parts where the gradient doesn't change a great deal.  This is what monokekie was talking about in relation to why the pH changes at different rates.  It depends on the concentration of hydronium ions present in the solution.

Thanks, does this mean that the rapid increase will always be constant? This is because the rule is -log[H3O+]

[stonecold]

I guess, but it really depends on the value of [H3O+].

What is the question?  Is this from Heinemann Chem 3/4?

[kenhung123]

Its not really a question just a concept I don't understand. The log graphs actually vary in shape and its a volume vs pH graph. I don't know why there is a rapid increase for a certain volume for different acid/bases

[stonecold]

It think it is due to the hydronium ion concertration.  i.e.  HNO₃ will have a different [H₃O⁺] concentration than H₂SO₄ for the same number of mol, however the [H₃O⁺] concentration when the number of moles of of HNO₃ is exactly 2 times the number of moles of H₂SO₄ will be the same.

e.g.  0.2 M HNO₃  has the same pH as 0.1 M H₂SO₄

Also note that as volume increases, the concentration decreases.

[kenhung123]

I think I got it guys. Basically, if you have an acid and a base titration, when you add small amount of strong acid slowly, the pH would change slowly at first because the strong base contains high OH- concentration. This requires a lot of H3O+ to bring it slightly down but when it has been brought down to pH7, just adding a little bit of H3O+ immediately adds onto the acidity of the solution because there is no more OH- to balance out. The H3O+ merely makes the solution more acidic as more is added. But once enough is added the pH change is less because the pH of the acid cannot go beyond itself.

Hope this helps for future people!

[stonecold]

That is pretty much much it Ken.  Just be careful not to confuse concentrated/dilute with strong/weak.  They mean completely different things.  Strength refers to the ability of the molecules in the solution to ionise completely or incompletely in water.  Concentration refers to the number of mol present in the solution (M or mol L^-1.)

[kenhung123]

The active ingredient in a brand of antacid is magnesium carbonate. A 1.30g tablet was crushed and placed in a conical flash. A small volume of water was added, the mixture stirred, and methyl orange indicator added. The indicator changed colour permanantly after 27.75mL of 1.00M hydrochloric acid has been added.

Write an equation for the reaction that occurs during the titration

Calculate the amount in mole of hydrochloric acid used to reach the end point

Calculate the percentage by mass of magnesium carbonate in the tablet Mr MgCO3-84

What is the function of the substances that make up the remaining mass of the tablet

Why do people often burp after taking an antacid tablet

[chem-nerd]

MgCO₃(aq) + 2HCl(aq) -> MgCl₂(aq) + CO₂(g) + H₂O(l)

n(HCl) = 0.02775 x 1.00

n(MgCO₃) = 1/2 n(HCl)

m(MgCO₃) = n(MgCO₃) x 84

use this to work out % by mass (m(MgCO₃)/1.30) X100

Other substances: binding agents, fillers, preservatives etc

Burping due to the CO₂ produced

[kenhung123]

Thank you!

A sample for air emitted from a car exhaust was taken and analysed for its sulfur dioxide (SO2) content. The gas sample was bubbled through 50ml of acidified 0.08M K2Cr2O7. The SO2 reacted with K2Cr2O7 to produce Cr 3+ and SO4 2-. The remaining amount of Cr2O7 2- was titrated with 0.2M SnCl2 solution.

Cr2O7 20 + 14H+ + 3Sn 2+ => 2Cr 3+ + 3Sn 4+ +7H2O

A titre of 10.2ml was required

The Redox equation is 3SO2+6H2O+Cr2O7 2- +2H+=> 3SO4 2- + 12H+ +2Cr 3+ + 7H2O

Find the amount of Cr2O7 2- in excess after reacting with SO2

Find the mass of SO2 present in the sample of air initially.

[Edmund]

Remaining amount of (using mole ratios from first equation):



Therefore the amount of in excess after the reaction is 0.00068mol


Total amount of at the start:



Amount reacted with :



Amount of required to react with (use mole ratios from redox equation):



Therefore the mass of :