another pka type question

[pizza1994]

A 2 M solution of the potent organic pollutant sodium methylthiolate (NaSCH3) is titrated to its endpoint with 2 M HCl. Which of the following best approximates the pH of the resultant solution? (pKb of SCH3– = 3.6)

answer is = -log ( 1* 10^-5.2)

this is from TPR

not sure what is happening here either.......

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[dnrs]

When you are at end point you have 2 M of HSCH3, which is the conjugate acid of your original base with pKa of 11.4. (pKa of the conjugate acid is 14 - pKb of the base)

Set up the equation of pKa = [H+][A-]/[HA] and you get 10^-11.4 = x^2 / 2, you get x = 1.4 * 10^-5.2, which is equal to your [H+].

To find pH, just take the -log of that.

 

[pizza1994]

When you are at end point you have 2 M of HSCH3, which is the conjugate acid of your original base with pKa of 11.4. (pKa of the conjugate acid is 14 - pKb of the base)

Set up the equation of pKa = [H+][A-]/[HA] and you get 10^-11.4 = x^2 / 2, you get x = 1.4 * 10^-5.2, which is equal to your [H+].

To find pH, just take the -log of that.

the 2 mol part is what i dont get......how the heck do you have 2 mol? this is been confusing me for so long!!!!!

 

[pizza1994]

the 2 mol part is what i dont get......how the heck do you have 2 mol? this is been confusing me for so long!!!!!

btw when can you use the eqn you provided? does it work all the time?

 

[E=mc^2]

the 2 mol part is what i dont get......how the heck do you have 2 mol? this is been confusing me for so long!!!!!
btw when can you use the eqn you provided? does it work all the time?

The two moles are because you added enough strong acid (HCl), 2 molar to be exact, to convert all of the base to conjugate acid. Therefore, your starting concentration of conjugate acid is 2 molar.

[HSCH3] <-> [H+]*[SCH3-] with a pka of 14-3.6= 10.4

So,

10^-10.4=x^2/[2M]

As for your second question, that equation is only valid when dealing with a weak base or a weak acid in equilibrium at a certain temperature. As you should know, strong acids and strong bases dissociate completely in aqueous solution.


EDIT- actually, I believe that it is not 2 molar, I beleive it is one molar. Because when you add a 2 molar equivalent to another 2 molar equivalent, you have 2Moles/liter +2 moles/liter of something else. The moles combined [H+] + [SCH3-] to get 2 mole of
[SCH3-] Meanwhile, you are also combining 1 liter+ 1 liter which will give you two liters of solution. Therefore your total concentration should be 1 molar solution giving you the equation,

10^-10.4=x^2/[1M]

Which would indeed add up to the answer you provided.

 

[pizza1994]

The two moles are because you added enough strong acid (HCl), 2 molar to be exact, to convert all of the base to conjugate acid. Therefore, your starting concentration of conjugate acid is 2 molar.

[HSCH3] <-> [H+]*[SCH3-] with a pka of 14-3.6= 10.4

So,

10^-10.4=x^2/[2M]

As for your second question, that equation is only valid when dealing with a weak base or a weak acid in equilibrium at a certain temperature. As you should know, strong acids and strong bases dissociate completely in aqueous solution.


EDIT- actually, I believe that it is not 2 molar, I beleive it is one molar. Because when you add a 2 molar equivalent to another 2 molar equivalent, you have 2Moles/liter +2 moles/liter of something else. The moles combined [H+] + [SCH3-] to get 2 mole of
[SCH3-] Meanwhile, you are also combining 1 liter+ 1 liter which will give you two liters of solution. Therefore your total concentration should be 1 molar solution giving you the equation,

10^-10.4=x^2/[1M]

Which would indeed add up to the answer you provided.

I still dont get how you got 1 molar. you add 2 M NaSCH3 to 2 M HCl but why is your SCH3- 1 M.......I know that C= n/V is the formula.....but when you add moles shouldnt you get 4 moles of SCH3-......I suck at this

 

[dnrs]

Because they combine to make 1 product.

 

[pizza1994]

I dont get it still. how does 2 M + 2M= 1 M?????

 

[pizza1994]

Because they combine to make 1 product.

Yeah but mathematically I dont see how you get 1M. because Im trying to use CV1 + CV2 = CV (final)

so... (2 M) (1 L) + (2 M) (1 L)= (C?) (2 L)

and then final Concentrtaion is 2M

 

[dnrs]

You are combining 2 moles of A and 2 moles of B to make 2 moles of AB.

 

[pizza1994]

You are combining 2 moles of A and 2 moles of B to make 2 moles of AB.

why isnt 2 moles of A + 2 moles of B= 4 moles of AB?

isnt it suppose to be additive?

 

[dnrs]

Think about it. Try using distributive property to help you.

 

[IslandStyle808]

why isnt 2 moles of A + 2 moles of B= 4 moles of AB?

isnt it suppose to be additive?

Well think backward from your product. If you were to dissociate 4 moles of AB, it would be 4 moles A and 4 moles B (not 8 moles A and 8 moles of B). So that can't be right it has to add up to 2 moles AB, since the reaction is associative.