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.
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.
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