Buffers
Calculate the pH of a solution formed when 50.0cm^3 of 0.120 moldm^-3 HCl was reacted with 50.0cm^3 of 0.280 moldm^-3 Sr(OH)2
mol HCl= 6 x 10^-3
mol Sr(OH)2= 0.014
2HCl + Sr(OH)2= SrCl2 + 2H2O
HCl is the limiting reagent, 3 x 10^-3 mol of SrCl2 are formed
concentration of SrCl2= 3 x 10^-3 mol/ 100/1000 to get into dm^3
Then how do I find the pH of the solution formed- answer is 13.34.
mol HCl= 6 x 10^-3
mol Sr(OH)2= 0.014
2HCl + Sr(OH)2= SrCl2 + 2H2O
HCl is the limiting reagent, 3 x 10^-3 mol of SrCl2 are formed
concentration of SrCl2= 3 x 10^-3 mol/ 100/1000 to get into dm^3
Then how do I find the pH of the solution formed- answer is 13.34.
Original post by ursbhhdb
Calculate the pH of a solution formed when 50.0cm^3 of 0.120 moldm^-3 HCl was reacted with 50.0cm^3 of 0.280 moldm^-3 Sr(OH)2
mol HCl= 6 x 10^-3
mol Sr(OH)2= 0.014
2HCl + Sr(OH)2= SrCl2 + 2H2O
HCl is the limiting reagent, 3 x 10^-3 mol of SrCl2 are formed
concentration of SrCl2= 3 x 10^-3 mol/ 100/1000 to get into dm^3
Then how do I find the pH of the solution formed- answer is 13.34.
mol HCl= 6 x 10^-3
mol Sr(OH)2= 0.014
2HCl + Sr(OH)2= SrCl2 + 2H2O
HCl is the limiting reagent, 3 x 10^-3 mol of SrCl2 are formed
concentration of SrCl2= 3 x 10^-3 mol/ 100/1000 to get into dm^3
Then how do I find the pH of the solution formed- answer is 13.34.
A couple of helpers for you.
1. Buffers only form when you have a weak acid mixed with its conjugate base (or vice versa with weak bases). This is not a buffer.
2. The SrCl2 is a salt and doesn't affect the pH.
3. Calculate the conc. of the OH- ions, which you've recognised as being in excess.
4. Use Kw or pKw (I much prefer pKw) to calculate the pH.
Original post by Pigster
A couple of helpers for you.
1. Buffers only form when you have a weak acid mixed with its conjugate base (or vice versa with weak bases). This is not a buffer.
2. The SrCl2 is a salt and doesn't affect the pH.
3. Calculate the conc. of the OH- ions, which you've recognised as being in excess.
4. Use Kw or pKw (I much prefer pKw) to calculate the pH.
1. Buffers only form when you have a weak acid mixed with its conjugate base (or vice versa with weak bases). This is not a buffer.
2. The SrCl2 is a salt and doesn't affect the pH.
3. Calculate the conc. of the OH- ions, which you've recognised as being in excess.
4. Use Kw or pKw (I much prefer pKw) to calculate the pH.
thanks, I'm still a little confused and can't get to the correct answer
I thought pH= -log (Kw/ [OH-)
pH= - log (1.00 x 10-14 / 0.014/100/1000)?
Original post by ursbhhdb
thanks, I'm still a little confused and can't get to the correct answer
I thought pH= -log (Kw/ [OH-)
pH= - log (1.00 x 10-14 / 0.014/100/1000)?
I thought pH= -log (Kw/ [OH-)
pH= - log (1.00 x 10-14 / 0.014/100/1000)?
n(OH-) has changed and hence [OH-], 1. because some it has reacted with H+ 2. volume has changed.
Find moles of HCl, and thus the moles of H+ ions.
Find moles of Sr(OH)2, and thus the moles of OH- ions which would be twice the amount as (OH)2.
Then find which reactant is in excess (has most moles). The moles of H+ or OH- that don’t react (neutralise) will cause the pH to change.
Convert this value into a concentration by dividing by the total volume of both solutions, and into dm^3. This will give you [OH-] or [H+],
Then it’s easy to work out pH. If you have [OH-] excess, then take pH = 14 - pOH, if you have [H+] then pH = -log[H+].
Find moles of Sr(OH)2, and thus the moles of OH- ions which would be twice the amount as (OH)2.
Then find which reactant is in excess (has most moles). The moles of H+ or OH- that don’t react (neutralise) will cause the pH to change.
Convert this value into a concentration by dividing by the total volume of both solutions, and into dm^3. This will give you [OH-] or [H+],
Then it’s easy to work out pH. If you have [OH-] excess, then take pH = 14 - pOH, if you have [H+] then pH = -log[H+].
(edited 4 years ago)
Original post by Pigster
n(OH-) has changed and hence [OH-], 1. because some it has reacted with H+ 2. volume has changed.
please may you do a step by step calculation?
Original post by Pigster
n(OH-) has changed and hence [OH-], 1. because some it has reacted with H 2. volume has changed.
So the total volume isn't 100cm3? ( with 100cm3 as total volume im getting pH as 13 tho)
(edited 4 years ago)
Original post by lyer_in_hellfyre
So the total volume isn't 100cm3?
It is 100.
My point was that since both the amount of OH- and the volume that the OH- is dissolved into have changed, then both need to be factored in.
Original post by Pigster
It is 100.
My point was that since both the amount of OH- and the volume that the OH- is dissolved into have changed, then both need to be factored in.
My point was that since both the amount of OH- and the volume that the OH- is dissolved into have changed, then both need to be factored in.
Ah ok my bad i misread
Original post by ursbhhdb
please may you do a step by step calculation?
Both Pigster's and Deggs_14's explanations are really good...if i may i ask, where did you get stuck?
Is it about subtracting the moles part to find [OH] concentration?
Imagine this :- A and B reacted to give C. At the end of reaction, we still have a part of B remaining. This unreacted moles of B's is used to find B's concentration we find to give us the pH.
So how to find the remaining moles? We know the total moles of B is 0.014 and we know how many moles of B reacted (3 x 10^-3) so...guess whats next?
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