chemistry bases help please
hi, please could I have some help on this question please? I'm using the hasselbach equation but I'm not sure how to work out the concentration of ACES?
question: https://ibb.co/4ncBJWG1
Thanks!
question: https://ibb.co/4ncBJWG1
Thanks!
Reply 1
Original post by anonymous56754
hi, please could I have some help on this question please? I'm using the hasselbach equation but I'm not sure how to work out the concentration of ACES?
question: https://ibb.co/4ncBJWG1
Thanks!
question: https://ibb.co/4ncBJWG1
Thanks!
a) : Molarity of ACES^- (conjugate base)
You start with 3.50 g of sodium ACES salt. The molar mass of ACES^- is 204.1 g/mol as per question text.
Doing the calculation: 3.50 g divided by 204.1 g/mol gives you about 0.01715 moles.
b) : Moles of HCl added
Next, let’s figure out how many moles of HCl you added. You have a concentration of 0.200 M and a volume of 50.0 mL, which is 0.0500 L.
Doing the calculation: 0.200 M times 0.0500 L equals 0.0100 moles.
c) : Reaction between ACES^- and HCl
When ACES^- reacts with HCl, it forms the weak acid HACES, reducing the amount of ACES^- and releasing H^+.
This means the moles of HACES formed equals the moles of HCl added, which is 0.0100 moles. So the moles of ACES^- left are 0.01715 - 0.0100, which is 0.00715 moles.
d) : Final solution concentrations
Now, let's find the concentrations of HACES and ACES^- in the final solution. They are ionizable, and the total volume is 250.0 mL, or 0.250 L.
For HACES, the concentration is 0.0100 moles divided by 0.250 L, which equals 0.0400 M. For ACES^-, it’s 0.00715 moles divided by 0.250 L, giving you 0.0286 M.
e) : Henderson-Hasselbalch Equation
Now, let's use the Henderson-Hasselbalch equation for HACES and ACES^-. You already know the pKa of ACES is 6.88.
You know: pH = pKa + log(HACES]/[ACES^-]) = 6.88 − 0.146, which is about 6.73.
So, the pH of the buffer solution is 6.73 (rounded to two decimal places).
Bye,
Sandro
(edited 1 month ago)
Reply 2
Original post by anonymous56754
hi, please could I have some help on this question please? I'm using the hasselbach equation but I'm not sure how to work out the concentration of ACES?
question: https://ibb.co/4ncBJWG1
Thanks!
question: https://ibb.co/4ncBJWG1
Thanks!
Personally, I wouldn’t use the HH equation as it’s annoying to memorise and it’s generally easier to use Ka = [H^+][A^-]/[HA].
In the question, they’ve told you that you are reacting sodium ACES with HCl, which makes ACES. You also know the reacting ratio is 1:1 and so the moles of the limiting reagent will correspond to the moles of ACES formed.
A big hint is that they’ve said that the solution made is a buffer solution. That means there must be an excess of the sodium ACES because a buffer solution must necessarily contain both the conjugate acid and the conjugate base - an excess of acid would ensure that all the conjugate base was used up and so the solution wouldn’t buffer. So HCl is the limiting reagent and hence the moles of HCl = the moles of ACES.
The moles of sodium ACES left should equal the original number of moles of sodium ACES - the moles of HCl.
(edited 1 month ago)
Reply 3
Original post by Nitrotoluene
a) : Molarity of ACES^- (conjugate base)
You start with 3.50 g of sodium ACES salt. The molar mass of ACES^- is 204.1 g/mol as per question text.
Doing the calculation: 3.50 g divided by 204.1 g/mol gives you about 0.01715 moles.
b) : Moles of HCl added
Next, let’s figure out how many moles of HCl you added. You have a concentration of 0.200 M and a volume of 50.0 mL, which is 0.0500 L.
Doing the calculation: 0.200 M times 0.0500 L equals 0.0100 moles.
c) : Reaction between ACES^- and HCl
When ACES^- reacts with HCl, it forms the weak acid HACES, reducing the amount of ACES^- and releasing H^+.
This means the moles of HACES formed equals the moles of HCl added, which is 0.0100 moles. So the moles of ACES^- left are 0.01715 - 0.0100, which is 0.00715 moles.
d) : Final solution concentrations
Now, let's find the concentrations of HACES and ACES^- in the final solution. They are ionizable, and the total volume is 250.0 mL, or 0.250 L.
For HACES, the concentration is 0.0100 moles divided by 0.250 L, which equals 0.0400 M. For ACES⁻, it’s 0.00715 moles divided by 0.250 L, giving you 0.0286 M.
e) : Henderson-Hasselbalch Equation
Now, let's use the Henderson-Hasselbalch equation for HACES and ACES^-. You already know the pKa of ACES is 6.88.
You know: pH = pKa + log([HACES]/[ACES^-]) = 6.88 − 0.146, which is about 6.73.
So, the pH of the buffer solution is 6.73 (rounded to two decimal places).
Bye,
Sandro
You start with 3.50 g of sodium ACES salt. The molar mass of ACES^- is 204.1 g/mol as per question text.
Doing the calculation: 3.50 g divided by 204.1 g/mol gives you about 0.01715 moles.
b) : Moles of HCl added
Next, let’s figure out how many moles of HCl you added. You have a concentration of 0.200 M and a volume of 50.0 mL, which is 0.0500 L.
Doing the calculation: 0.200 M times 0.0500 L equals 0.0100 moles.
c) : Reaction between ACES^- and HCl
When ACES^- reacts with HCl, it forms the weak acid HACES, reducing the amount of ACES^- and releasing H^+.
This means the moles of HACES formed equals the moles of HCl added, which is 0.0100 moles. So the moles of ACES^- left are 0.01715 - 0.0100, which is 0.00715 moles.
d) : Final solution concentrations
Now, let's find the concentrations of HACES and ACES^- in the final solution. They are ionizable, and the total volume is 250.0 mL, or 0.250 L.
For HACES, the concentration is 0.0100 moles divided by 0.250 L, which equals 0.0400 M. For ACES⁻, it’s 0.00715 moles divided by 0.250 L, giving you 0.0286 M.
e) : Henderson-Hasselbalch Equation
Now, let's use the Henderson-Hasselbalch equation for HACES and ACES^-. You already know the pKa of ACES is 6.88.
You know: pH = pKa + log([HACES]/[ACES^-]) = 6.88 − 0.146, which is about 6.73.
So, the pH of the buffer solution is 6.73 (rounded to two decimal places).
Bye,
Sandro
That’s correct thanks!
Reply 4
Original post by TypicalNerd
Personally, I wouldn’t use the HH equation as it’s annoying to memorise and it’s generally easier to use Ka = [H^+][A^-]/[HA].
In the question, they’ve told you that you are reacting sodium ACES with HCl, which makes ACES. You also know the reacting ratio is 1:1 and so the moles of the limiting reagent will correspond to the moles of ACES formed.
A big hint is that they’ve said that the solution made is a buffer solution. That means there must be an excess of the sodium ACES because a buffer solution must necessarily contain both the conjugate acid and the conjugate base - an excess of acid would ensure that all the conjugate base was used up and so the solution wouldn’t buffer. So HCl is the limiting reagent and hence the moles of HCl = the moles of ACES.
The moles of sodium ACES left should equal the original number of moles of sodium ACES - the moles of HCl.
In the question, they’ve told you that you are reacting sodium ACES with HCl, which makes ACES. You also know the reacting ratio is 1:1 and so the moles of the limiting reagent will correspond to the moles of ACES formed.
A big hint is that they’ve said that the solution made is a buffer solution. That means there must be an excess of the sodium ACES because a buffer solution must necessarily contain both the conjugate acid and the conjugate base - an excess of acid would ensure that all the conjugate base was used up and so the solution wouldn’t buffer. So HCl is the limiting reagent and hence the moles of HCl = the moles of ACES.
The moles of sodium ACES left should equal the original number of moles of sodium ACES - the moles of HCl.
How do you know the sodium salt reacts with HCl to form aces?
Reply 5
Original post by anonymous56754
How do you know the sodium salt reacts with HCl to form aces?
When a salt of a weak acid reacts with a stronger acid, you always make the weak acid and the salt of the stronger acid. HCl is a strong acid (as you need to memorise) and ACES is a weak acid (as it tells you in the question), so you can say with certainty that sodium ACES will protonate to ACES when HCl is added
Reply 6
Original post by TypicalNerd
When a salt of a weak acid reacts with a stronger acid, you always make the weak acid and the salt of the stronger acid. HCl is a strong acid (as you need to memorise) and ACES is a weak acid (as it tells you in the question), so you can say with certainty that sodium ACES will protonate to ACES when HCl is added
Ohh, ok thank you!
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