Calculate the number of moles of NaOBr(s) that would have to be added to 125 mL of 0.160
M HOBr to produce a buffer solution with [H+] = 5.00 10–9 M. Assume that volume change
is negligible.
The first step in solving this problem is to write out the balanced chemical equation for the reaction between NaOBr and HOBr:
NaOBr + HOBr → NaBr + H2O + O2
Next, we need to determine the pH of the buffer solution. In order to do this, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A–]/[HA])
where pKa is the dissociation constant of HOBr and [A–]/[HA] is the ratio of the concentrations of the conjugate base (OBr–) to the acid (HOBr) in the buffer solution. Since the pH is given as 5.00 × 10–9, the pOH is 14 – 5.00 × 10–9 = 13.9999995. We can convert this to [OH–] using the relationship:
pOH = –log([OH–])
[OH–] = 10–pOH = 3.162 × 10–14 M
The dissociation constant of HOBr can be found in a table of acid dissociation constants or calculated using the equilibrium expression:
Ka = [H+][OBr–]/[HOBr]
At pH = 5.00 × 10–9, [H+] = 5.00 × 10–9 M and [OBr–]/[HOBr] is equal to the conjugate base-to-acid ratio:
[A–]/[HA] = [OBr–]/[HOBr] = 10–pKa
Solving for pKa, we get:
pKa = –log(Ka) = –log([H+][OBr–]/[HOBr]) = –log(5.00 × 10–9 × 10–13.9999995/0.160) = 8.5123
Now we can use the Henderson-Hasselbalch equation to solve for [A–]/[HA]:
5.00 × 10–9 = 8.5123 + log([OBr–]/[HOBr])
[OBr–]/[HOBr] = antilog(5.00 × 10–9 – 8.5123) = 4.556 × 10–4
Since the volume change is negligible, we can assume that the final volume of the buffer solution is still 125 mL. Therefore, the concentration of HOBr must be equal to the concentration of the NaOBr that is added to the solution. We can use the following equation to relate moles of NaOBr to its concentration:
moles NaOBr = concentration NaOBr × volume NaOBr
where the volume of NaOBr is the volume needed to make the total volume of the buffer solution up to 125 mL.
The moles of HOBr in 125 mL of 0.160 M solution is:
moles HOBr = concentration HOBr × volume HOBr = 0.160 × 0.125 = 0.0200 moles
Since the ratio of [OBr–]/[HOBr] is 4.556 × 10–4, the concentration of NaOBr must be:
concentration NaOBr = concentration HOBr/[OBr–]/[HOBr] = 0.160/4.556 × 10–4 = 351.5 M
The number of moles of NaOBr needed to make 125 mL of 351.5 M solution is:
moles NaOBr = concentration NaOBr × volume NaOBr = 351.5 × (125 – 0.125) × 10–3 = 43.9 moles
Therefore, 43.9 moles of NaOBr would have to be added to 125 mL of 0.160 M HOBr to produce a buffer solution with [H+] = 5.00 × 10–9 M.
To calculate the number of moles of NaOBr required to produce a buffer solution with a specific [H+], we need to use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
In this case, we have [H+] = 5.00 × 10^–9 M and we assume the buffer components are HOBr and OBr– (from NaOBr). The pKa of HOBr is required to calculate the values for [HA] and [A-].
1. Calculate the pKa of HOBr:
Given that pKa = -log(Ka), where Ka is the acid dissociation constant, we need to find the Ka of HOBr. Because the Ka of HOBr is not given, we must use the relationship:
pKa + pKb = 14
Assuming that Kb for the OBr- ion is small in comparison to the water auto-ionization constant, we can approximate Kb to be negligible. Therefore, we can calculate the pKa as:
pKa = 14 - pKb
Given that pKb of water is 14, we can substitute that value:
pKa = 14 - 14 = 0
Thus, the pKa of HOBr is 0.
2. Convert the pH of [H+] to pOH.
Using the relationship:
pH + pOH = 14
We can determine the pOH value as:
pOH = 14 - pH = 14 - (-log([H+])) = 14 - (-log(5.00 × 10^–9)) = 14 + 9 = 23
3. Calculate the concentration of [A-] from pOH.
Using the relationship:
pOH = -log([OH-])
We can rewrite it as:
[OH-] = 10^(-pOH)
Substituting the pOH value:
[OH-] = 10^(-23)
Since OBr- is the conjugate base of HOBr, [A-] = [OH-].
4. Calculate [HA] from [H+].
Using the relationship:
[HA] = [H+] / (10^(pKa - pH))
Substituting the values:
[HA] = 5.00 × 10^–9 / (10^(0 - 14)) = 5.00 × 10^–9 / 10^(-14) = 5.00 × 10^–9 / 10^14 = 5.00 × 10^5
5. Calculate the moles of HOBr required.
The volume is given as 125 mL, and the molarity is given as 0.160 M.
moles of HOBr = volume (L) x molarity (M)
= 0.125 L x 0.160 M
= 0.02 moles
6. Calculate moles of NaOBr required.
Since we assume that the volume change is negligible, the moles of NaOBr required will be the same as the moles of HOBr required.
Therefore, the number of moles of NaOBr(s) that would have to be added to 125 mL of 0.160 M HOBr to produce a buffer solution with [H+] = 5.00 × 10–9 M is 0.02 moles.