Calculating pH of a Na2HPO4 solution...

[spikers220]

I dont know why I can't the right answer but any help would be great!

Calculate the pH of the solution containing 3.875g of Na2HPO4 (MW. = 141.98 g/mol) that has been dissolved in a 250 mL volumetric flask and diluted to the mark.


The three equilibria values for H3PO4, H2PO4-, and HPO42- are:
Ka1 = 7.11 x 10 ^ -3
Ka2 = 6.34 x 10 ^ -8
Ka3 = 4.22 x 10 ^ -11

I get pH = 5.67, but that is incorrect. I am solving using just the final equilibria value but I must have to take into account other values also somehow? But if I do...I dont know how?

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[rockclock]

I dont know why I can't the right answer but any help would be great!

Calculate the pH of the solution containing 3.875g of Na2HPO4 (MW. = 141.98 g/mol) that has been dissolved in a 250 mL volumetric flask and diluted to the mark.


The three equilibria values for H3PO4, H2PO4-, and HPO42- are:
Ka1 = 7.11 x 10 ^ -3
Ka2 = 6.34 x 10 ^ -8
Ka3 = 4.22 x 10 ^ -11

I get pH = 5.67, but that is incorrect. I am solving using just the final equilibria value but I must have to take into account other values also somehow? But if I do...I dont know how?

that very small Ka3 value tells you HPO4^2- is a weak acid.

using Ka3 assumes your anion will deprotonate further, which is not the case. it's actually going to accept a proton so you will need to calculate the Kb for the reverse of Ka2, and go from there as a base problem. try that and let me know if it works. i'll try it myself with some scratch paper when i get a chance.

 

[rockclock]

double check my work...i just did this on a napkin and wolfram...

Kb = 1.58e-7
[OH-] = 0.000131
pOH = 3.88
pH = 10.12

 

[spoog74]

double check my work...i just did this on a napkin and wolfram...

Kb = 1.58e-7
[OH-] = 0.000131
pOH = 3.88
pH = 10.12

can you explain how you got those and why? Whats the exact reasoning behind using ka value of only 2nd one and not first or third... Please

Thanks

 

[rockclock]

haha you're lucky i'm bored out of my mind at work...haven't typed up a solution like that in a long time...

 

[Josh779]

Find the molarity first. Then divide 1x10^-14/Ka2. This will give you the Kb.
Next use this equation [OH-]=SQ KbxMolarity
Next take the -log of [OH-] which gives you the pOH
Lastly subtract from 14 to find the pH.

Edit: ^^ that works too, haha!

 

[rockclock]

Find the molarity first. Then divide 1x10^-14/Ka2. This will give you the Kb.
Next use this equation [OH-]=SQ KbxMolarity
Next take the -log of [OH-] which gives you the pOH
Lastly subtract from 14 to find the pH.

Edit: ^^ that works too, haha!

haha we're completely on the same page homie...just tried to spell it out a little bit

 

[spoog74]

haha you're lucky i'm bored out of my mind at work...haven't typed up a solution like that in a long time...

im sorry to be a bit slow on this, but why couldnt you use k1 for this problem? Is it because it wont dissociate to form either a base or an acid?

 

[rockclock]

im sorry to be a bit slow on this, but why couldnt you use k1 for this problem? Is it because it wont dissociate to form either a base or an acid?

you're putting HPO4^2- into water. write out what the equation for ka1 would be...it doesn't have HPO4^2- as either a reactant or a product.

no matter how many ka's they give you, you're only going to ever be choosing between two...whatever you're putting into solution can either act as an acid (donate a proton) or act as a base (accept a proton)...those are your only two choices. in this case that's either 3a (acid) or 2b (base). 2b wins because kb2 > ka3

 

[Golfguy]

Gents, what is "ICE" and SQ?

 

[rockclock]

SQ is square root...that's basically a shortcut formula to what you would get if you did the ICE box.

ICE box is a table that list the Initial concentration, Change in concentration, and End concentration for all reactants and products in a reaction. In the 2b reaction, for example (you ignore [H2O] since it is a pure liquid):

Initially:
[HPO4^2-] = what you're given in the problem, convert grams to moles, divide by volume
[each product] = 0

Change:
[HPO4^2-] decreases by X, depending on how many "net" molecules react to reach equilibrium
[each product] increases by X

In the end, your Kb2 expression would be

kb2 = [X]^2/[original HPO4 conc. - X]

since X is going to be orders of magnitude smaller than the original HPO4 conc., you can just drop it in the denominator for easy solving. it reduces to that SQRT formula mentioned earlier:

kb2 = [X]^2/[original HPO4 conc.]

plug in everything, solve for X, which will give you the final [OH-]. from there, pOH, and then pH.

 

[Josh779]

Not sure about the ice thing, but SQ means square root.


Edit: ^^ what he said, haha

 

[yappy]

Rock - if you dont mind me asking what was your DAT score? You seem very solid on your sciences! I wish I had you as a personal tutor.

SQ is square root...that's basically a shortcut formula to what you would get if you did the ICE box.

ICE box is a table that list the Initial concentration, Change in concentration, and End concentration for all reactants and products in a reaction. In the 2b reaction, for example (you ignore [H2O] since it is a pure liquid):

Initially:
[HPO4^2-] = what you're given in the problem, convert grams to moles, divide by volume
[each product] = 0

Change:
[HPO4^2-] decreases by X, depending on how many "net" molecules react to reach equilibrium
[each product] increases by X

In the end, your Kb2 expression would be

kb2 = [X]^2/[original HPO4 conc. - X]

since X is going to be orders of magnitude smaller than the original HPO4 conc., you can just drop it in the denominator for easy solving. it reduces to that SQRT formula mentioned earlier:

kb2 = [X]^2/[original HPO4 conc.]

plug in everything, solve for X, which will give you the final [OH-]. from there, pOH, and then pH.

 

[Josh779]

bump

 

[spikers220]

Thanks a ton guys.. I dont know why I was thinking it was going to loose that other hydrogen to become PO43-... Got it now!

 

[rockclock]

Thanks a ton guys.. I dont know why I was thinking it was going to loose that other hydrogen to become PO43-... Got it now!

did you know that the right answer was 10... all along? you must have since you knew your 5... was wrong, right? in these problems, being on the wrong side of neutral is the red flag that you picked the wrong direction (gaining/losing a proton)

 

[rockclock]

Rock - if you dont mind me asking what was your DAT score? You seem very solid on your sciences! I wish I had you as a personal tutor.

haha i held my own...still don't know if taking it the monday after finals was the best move though.

 

[spikers220]

did you know that the right answer was 10... all along? you must have since you knew your 5... was wrong, right? in these problems, being on the wrong side of neutral is the red flag that you picked the wrong direction (gaining/losing a proton)

Thanks for the advice. I don't know why but I just had it in my head that it was going to loose that last hydrogen.