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
(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.
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.
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.
To solve this problem, we need to use the balanced chemical equation between Ba(OH)2 and HOBr to determine the stoichiometric ratio and then use the concentration and volume of the HOBr solution to calculate the volume of Ba(OH)2 needed.
The balanced equation for the reaction between Ba(OH)2 and HOBr is:
Ba(OH)2 + 2HOBr -> Ba(Br)2 + 2H2O
(i) To calculate the volume of Ba(OH)2 solution needed to reach the equivalence point, we need to determine the number of moles of HOBr present in the 65.0 mL solution of 0.146 M HOBr.
Number of moles of HOBr = concentration * volume
= 0.146 M * 0.0650 L
= 0.00949 moles
From the balanced equation, we can see that 1 mole of Ba(OH)2 reacts with 2 moles of HOBr. Therefore, the number of moles of Ba(OH)2 needed is half the number of moles of HOBr.
Number of moles of Ba(OH)2 = 0.00949 moles / 2
= 0.00474 moles
Next, we can use the concentration and volume of the Ba(OH)2 solution to calculate the volume needed.
Number of moles of Ba(OH)2 = concentration * volume
Volume = Number of moles of Ba(OH)2 / concentration
Volume = 0.00474 moles / 0.115 M
= 0.0412 L
= 41.2 mL
Therefore, the volume of 0.115 M Ba(OH)2 solution needed to reach the equivalence point is 41.2 mL.
(ii) At the equivalence point, the reaction between Ba(OH)2 and HOBr is complete. Ba(OH)2 is a strong base while HOBr is a weak acid. The reaction produces Ba(Br)2, which is a neutral salt, and water (H2O).
Since a strong base and a weak acid have reacted, the resulting solution at the equivalence point will have a pH greater than 7. The water formed will not affect the pH significantly. Hence, the pH at the equivalence point is greater than 7.
To find 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), we can use the concept of stoichiometry and balanced chemical equations.
(i) The balanced chemical equation for the reaction between Ba(OH)2 and HOBr is:
Ba(OH)2(aq) + 2 HOBr(aq) → Ba(Br)2(aq) + 2 H2O(l)
From the balanced equation, we can determine the mole ratio between Ba(OH)2 and HOBr, which is 1:2. This means that for every mole of Ba(OH)2 reacted, 2 moles of HOBr are consumed.
First, let's calculate the number of moles of HOBr in the 65.0 mL solution using the formula:
moles = concentration × volume (in liters)
moles of HOBr = 0.146 M × 0.065 L
= 0.00949 moles
Since the mole ratio between Ba(OH)2 and HOBr is 1:2, we need half the number of moles of Ba(OH)2 to react with the HOBr. Thus, the number of moles of Ba(OH)2 required is:
moles of Ba(OH)2 = 0.00949 moles / 2
= 0.00475 moles
To find the volume of 0.115 M Ba(OH)2(aq), we can rearrange the formula:
moles = concentration × volume (in liters)
Volume of Ba(OH)2(aq) = moles / concentration
= 0.00475 moles / 0.115 M
= 0.0413 L
Finally, convert the volume from liters to milliliters:
Volume of Ba(OH)2(aq) = 0.0413 L × 1000 mL/L
= 41.3 mL
Therefore, the volume of 0.115 M Ba(OH)2(aq) needed to reach the equivalence point is approximately 41.3 mL.
(ii) At the equivalence point, the reaction is complete, and all the HOBr has reacted with Ba(OH)2. The products formed are Ba(Br)2(aq) (barium bromide) and H2O(l) (water). Barium bromide is a strong electrolyte and dissociates completely into ions in solution.
Since bromide ions (Br-) are anions, they will not affect the pH directly. However, water (H2O) can undergo autoionization, resulting in the formation of hydronium ions (H3O+) and hydroxide ions (OH-):
H2O(l) ↔ H3O+(aq) + OH-(aq)
At neutral pH, the concentration of hydronium ions (H3O+) is equal to the concentration of hydroxide ions (OH-), resulting in a pH of 7. Since the reaction is complete at the equivalence point and only the products of the reaction are present, the pH at the equivalence point of this titration will be equal to 7.