(i) The balanced chemical equation for the reaction between Ba(OH)2 and HOBr is:
Ba(OH)2 + 2HOBr → Ba(Br)2 + 2H2O
From the equation, we see that the mole ratio of Ba(OH)2 to HOBr is 1:2. This means that 1 mole of Ba(OH)2 will react with 2 moles of HOBr at the equivalence point. To determine the volume of Ba(OH)2 needed, we can use the following equation:
(0.115 mol/L) x (V L) = (2 x 0.146 mol/L) x (0.0650 L)
where V is the volume of Ba(OH)2 needed to reach the equivalence point. Solving for V, we get:
V = (2 x 0.146 mol/L x 0.0650 L) / 0.115 mol/L = 0.203 L
Therefore, the volume of 0.115 M Ba(OH)2(aq) needed to reach the equivalence point when titrated into a 65.0 mL sample of 0.146 M HOBr(aq) is 0.203 L (or 203 mL).
(ii) The pH at the equivalence point depends on the nature of the acid and base involved in the reaction. HOBr is a weak acid with a pKa of 8.7, while Ba(OH)2 is a strong base. At the equivalence point, all of the HOBr will have been neutralized by the Ba(OH)2, giving a solution of Ba(Br)2 and water.
Ba(OH)2 + 2HOBr → Ba(Br)2 + 2H2O
The pH of the solution at the equivalence point will depend on the salt formed. Ba(Br)2 is a salt of a strong base (Ba(OH)2) and a weak acid (HBr). It will therefore hydrolyze to a small extent, producing Br- ions and some OH- ions. The OH- ions will react with any remaining H+ ions from the HOBr that was not completely neutralized at the equivalence point, resulting in a slightly basic solution. Therefore, the pH at the equivalence point will be slightly greater than 7.
A solution of Ba(OH)2 is titrated into a solution of HOBг.
(i) Calculate the volume of 0.115 M Ba(OH)2(aq) needed to reach the equivalence point when
titrated into a 65.0 mL sample of 0.146 M HOBr(aq).
(ii) Indicate whether the pH at the equivalence point is less than 7, equal to 7, or greater than 7
3 answers
Calculate the number of moles of NaOBrs) that would have to be added to 125 mL of 0.160
M HOBr to produce a buffer solution with [H^ + ]= 5 * 10 ^ - 9 * M . Assume that volume change
is negligible.
M HOBr to produce a buffer solution with [H^ + ]= 5 * 10 ^ - 9 * M . Assume that volume change
is negligible.
To create a buffer solution with a specific pH, we need a weak acid and its conjugate base (or a weak base and its conjugate acid) in roughly equal proportions. In this case, the weak acid is HOBr and its conjugate base is OBr-. The balanced chemical equation for the dissociation of HOBr is:
HOBr + H2O ⇌ H3O+ + OBr-
The Ka for HOBr is 2.3 × 10^-9, so the pKa is 8.64. At a pH of 8.3, the concentration of H3O+ is:
[H3O+] = 5 × 10^-9 M
Since the solution is at equilibrium, we can use the equilibrium expression for Ka to calculate the concentration of the conjugate base:
Ka = [H3O+][OBr-] / [HOBr]
2.3 × 10^-9 = (5 × 10^-9)([OBr-]) / (0.160 - [OBr-])
Solving for [OBr-], we get:
[OBr-] = 0.120 M
This means that the concentration of HOBr is:
[HOBr] = 0.160 - [OBr-] = 0.040 M
To create a buffer solution with these concentrations, we need to add NaOBr to the HOBr solution. The NaOBr will dissociate into Na+ and OBr-. Since we need the same concentration of OBr-, we need to add 0.120 M NaOBr. The volume of NaOBr solution needed can be calculated using the following equation:
moles of NaOBr = concentration of NaOBr × volume of NaOBr
Since we need 0.120 moles of OBr-, we need half as many moles of NaOBr (since the mole ratio of NaOBr to OBr- is 1:1):
moles of NaOBr = 0.060 moles
The concentration of NaOBr is not given, so we cannot directly calculate the volume needed. However, we can use the fact that 0.060 moles of NaOBr is equivalent to 0.120 moles of OBr- to calculate the volume indirectly. If the concentration of NaOBr is x M, then:
moles of NaOBr = concentration of NaOBr × volume of NaOBr
0.060 = x × volume of NaOBr
volume of NaOBr = 0.060 / x
We know that the total volume of the final solution is 125 mL, so:
volume of NaOBr + 125 mL = total volume
volume of NaOBr = total volume - 125 mL
Substituting this into the previous equation, we get:
0.060 / x = total volume - 125 mL
Solving for x, we get:
x = 0.24 M
Therefore, we need to add 0.60 moles of NaOBr to 125 mL of 0.160 M HOBr to create a buffer solution with a pH of 8.3. However, since the volume change is negligible, the final volume of the solution will be slightly larger than 125 mL.
HOBr + H2O ⇌ H3O+ + OBr-
The Ka for HOBr is 2.3 × 10^-9, so the pKa is 8.64. At a pH of 8.3, the concentration of H3O+ is:
[H3O+] = 5 × 10^-9 M
Since the solution is at equilibrium, we can use the equilibrium expression for Ka to calculate the concentration of the conjugate base:
Ka = [H3O+][OBr-] / [HOBr]
2.3 × 10^-9 = (5 × 10^-9)([OBr-]) / (0.160 - [OBr-])
Solving for [OBr-], we get:
[OBr-] = 0.120 M
This means that the concentration of HOBr is:
[HOBr] = 0.160 - [OBr-] = 0.040 M
To create a buffer solution with these concentrations, we need to add NaOBr to the HOBr solution. The NaOBr will dissociate into Na+ and OBr-. Since we need the same concentration of OBr-, we need to add 0.120 M NaOBr. The volume of NaOBr solution needed can be calculated using the following equation:
moles of NaOBr = concentration of NaOBr × volume of NaOBr
Since we need 0.120 moles of OBr-, we need half as many moles of NaOBr (since the mole ratio of NaOBr to OBr- is 1:1):
moles of NaOBr = 0.060 moles
The concentration of NaOBr is not given, so we cannot directly calculate the volume needed. However, we can use the fact that 0.060 moles of NaOBr is equivalent to 0.120 moles of OBr- to calculate the volume indirectly. If the concentration of NaOBr is x M, then:
moles of NaOBr = concentration of NaOBr × volume of NaOBr
0.060 = x × volume of NaOBr
volume of NaOBr = 0.060 / x
We know that the total volume of the final solution is 125 mL, so:
volume of NaOBr + 125 mL = total volume
volume of NaOBr = total volume - 125 mL
Substituting this into the previous equation, we get:
0.060 / x = total volume - 125 mL
Solving for x, we get:
x = 0.24 M
Therefore, we need to add 0.60 moles of NaOBr to 125 mL of 0.160 M HOBr to create a buffer solution with a pH of 8.3. However, since the volume change is negligible, the final volume of the solution will be slightly larger than 125 mL.
1 answer