Back
to top
Alevel Chemistry Exam Question on Buffers..help for geniuses
Watch this threadPage 1 of 1
Go to first unread
Skip to page:
GenuineHpLaptop
Badges:
9
Rep:
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#1
Okay so this is an OCR Alevel Chem question ...I dont understand how and why the question is tackled. Help please??
H2PO4- (minus) is a weak acid and has the Ka value of 6.3 x 10^-8 mol dm-3. For the reaction it loses one proton. Write the equation for the reaction and express the Ka (1m).
Calculate the pH of a 0.010moldm-3 solution of NaH2PO4 . (2m)
A mixture of NaH2PO4 and Na2HPO4 acts as a buffer in fertilisers. Calculate the pH of a solution containing equal amounts of NaH2PO4 and Na2HPO4. (1m)
the answer is -log(ka (6.3x 10^-8))... why so though? why is pka the pH? can someone explain this probably really simple question please
Okay so this is the main toughie? lol its probably not .. help please?
Some students wish to make a buffer solution of pH 7 using NaH2PO4 and Na2HPO4. Suggest suitable masses of each solid to dissolve into 1.0dm^3 of water. (4m)
H2PO4- (minus) is a weak acid and has the Ka value of 6.3 x 10^-8 mol dm-3. For the reaction it loses one proton. Write the equation for the reaction and express the Ka (1m).
Calculate the pH of a 0.010moldm-3 solution of NaH2PO4 . (2m)
A mixture of NaH2PO4 and Na2HPO4 acts as a buffer in fertilisers. Calculate the pH of a solution containing equal amounts of NaH2PO4 and Na2HPO4. (1m)
the answer is -log(ka (6.3x 10^-8))... why so though? why is pka the pH? can someone explain this probably really simple question please
Okay so this is the main toughie? lol its probably not .. help please?
Some students wish to make a buffer solution of pH 7 using NaH2PO4 and Na2HPO4. Suggest suitable masses of each solid to dissolve into 1.0dm^3 of water. (4m)
0
reply
M0nkey Thunder
Badges:
13
Rep:
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#2
Report
#2
You need to write the equation for the acid dissociating.
Rearrange the Ka equation to calculate the concentration of H+ ions to then calculate the pH from.
Where the concentration of weak acid and soluble salt of that acid are the same, you've reached the half-neutralisation point. The acid has dissociated halfway and so: Ka= [H+]. As a result pKa= pH.
For that last question, I believe you just need to figure out the ratio of the weak acid to its soluble salt. You already know the Ka value, can calculate the H+ concentration that you would have if the pH were to be 7. Just rearrange the Ka equation for that weak acid, find the ratio of [HA] and [A-] and you should be able to figure out suitable masses for both.
Rearrange the Ka equation to calculate the concentration of H+ ions to then calculate the pH from.
Where the concentration of weak acid and soluble salt of that acid are the same, you've reached the half-neutralisation point. The acid has dissociated halfway and so: Ka= [H+]. As a result pKa= pH.
For that last question, I believe you just need to figure out the ratio of the weak acid to its soluble salt. You already know the Ka value, can calculate the H+ concentration that you would have if the pH were to be 7. Just rearrange the Ka equation for that weak acid, find the ratio of [HA] and [A-] and you should be able to figure out suitable masses for both.
1
reply
Diana :)
Badges:
2
Rep:
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#3
GenuineHpLaptop
Badges:
9
Rep:
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#4
(Original post by M0nkey Thunder)
You need to write the equation for the acid dissociating.
Rearrange the Ka equation to calculate the concentration of H+ ions to then calculate the pH from.
Where the concentration of weak acid and soluble salt of that acid are the same, you've reached the half-neutralisation point. The acid has dissociated halfway and so: Ka= [H+]. As a result pKa= pH.
For that last question, I believe you just need to figure out the ratio of the weak acid to its soluble salt. You already know the Ka value, can calculate the H+ concentration that you would have if the pH were to be 7. Just rearrange the Ka equation for that weak acid, find the ratio of [HA] and [A-] and you should be able to figure out suitable masses for both.
You need to write the equation for the acid dissociating.
Rearrange the Ka equation to calculate the concentration of H+ ions to then calculate the pH from.
Where the concentration of weak acid and soluble salt of that acid are the same, you've reached the half-neutralisation point. The acid has dissociated halfway and so: Ka= [H+]. As a result pKa= pH.
For that last question, I believe you just need to figure out the ratio of the weak acid to its soluble salt. You already know the Ka value, can calculate the H+ concentration that you would have if the pH were to be 7. Just rearrange the Ka equation for that weak acid, find the ratio of [HA] and [A-] and you should be able to figure out suitable masses for both.
the answer for the 4 marker is put in the ratio 1.34 NaH2PO4 : 1 Na2HPO4 btw to anyone who works the answer?
)
0
reply
GenuineHpLaptop
Badges:
9
Rep:
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#5
(Original post by Diana :))
what paper is this question from?
what paper is this question from?
)
0
reply
Diana :)
Badges:
2
Rep:
?
You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#6
X
Page 1 of 1
Go to first unread
Skip to page:
Quick Reply
