Hypobromous acid, HOBr, is a weak acid that dissociates in water, as represented by the equation
below:
HOBrag) Haq) + OBrag
K.-2.3 x 10-9
(a) Calculate the value of [H] in an HOBr solution that has a pH of 4.95. (b) Write the equilibrium constant expression for the ionization of HOBr in water, then calculate
the concentration of HOBrag) in an HOBr solution that has [H] equal to 1.8 x 10 M.
(c) A solution of Ba(OH)2 is titrated into a solution of HOBr. (i) Calculate the volume of 0.115 M Ba(OH) needed to reach the equivalence point when
titrated into a 65.0 mL sample of 0.146 M HOB (ii) Indicate whether the pH at the equivalence point is less than 7, equal to 7, or greater than 7. Explain.
(d) Calculate the number of moles of NaOBr) that would have to be added to 125 mL of 0.160
MHOBT to produce a buffer solution with [H]=5.00 x 10-9 M. Assume that volume change is negligible.
(a) We can use the pH formula to find [H+]:
pH = -log[H+]
4.95 = -log[H+]
[H+] = 7.01 x 10^-5 M
(b) The equilibrium constant expression is:
K = [H+][OBr-]/[HOBr]
At equilibrium, [HOBr] = initial concentration - [H+] = 1.8 x 10^-5 M
Then, using the dissociation equation:
K = [H+][OBr-]/(1.8 x 10^-5 M)
2.3 x 10^-9 = [H+][OBr-]/(1.8 x 10^-5 M)
[OBr-] = 1.77 x 10^-12 M
(c)
(i) The balanced equation for the titration is:
Ba(OH)2 + 2HOBr → Ba(Br)2 + 2H2O
From the equation, we see that 2 moles of HOBr react with 1 mole of Ba(OH)2 at the equivalence point.
So the number of moles of HOBr in the solution at the equivalence point is:
n(HOBr) = M x V = 0.146 M x 65.0 mL / 1000 mL = 0.00949 mol
To react with all of the HOBr at the equivalence point, we need:
n(Ba(OH)2) = 2 x n(HOBr) = 0.01898 mol
The volume of 0.115 M Ba(OH)2 needed to contain this amount of moles is:
V = n / M = 0.01898 mol / 0.115 mol/L = 0.165 L = 165 mL
(ii) At the equivalence point, all of the HOBr has reacted with Ba(OH)2 to form the salt Ba(Br)2, which is a neutral salt. Therefore, the pH at the equivalence point is equal to 7.
(d) To make a buffer solution that has [H+] = 5.00 x 10^-9 M, we need to have equal amounts of HOBr and OBr- ions.
First, let's find the initial amount of moles of HOBr in the solution:
n(HOBr) = M x V = 0.160 M x 125 mL / 1000 mL = 0.02 mol
Since we need equal amount of HOBr and OBr-, we need to add 0.01 mol of NaOBr.
The concentration of NaOBr needed is:
0.01 mol / 0.125 L = 0.08 M
(a) To calculate the value of [H+] in an HOBr solution that has a pH of 4.95, we can use the formula pH = -log[H+]. Rearranging the formula, we have [H+] = 10^(-pH). Substituting the given pH value:
[H+] = 10^(-4.95)
= 7.08 x 10^(-5) M
Therefore, the concentration of [H+] in the HOBr solution is 7.08 x 10^(-5) M.
(b) The equilibrium constant expression for the ionization of HOBr in water is given as:
K = [H+][OBr-]/[HOBr]
Since the value of the equilibrium constant (K) is given as 2.3 x 10^(-9), we can rearrange the equation to solve for [HOBr]:
[HOBr] = [H+][OBr-]/K
Given that [H+] = 1.8 x 10^(-1) M, and assuming that the dissociation of HOBr is negligible compared to the concentration of [H+], we can substitute the values into the equation:
[HOBr] ≈ (1.8 x 10^(-1) M)(1.8 x 10^(-1) M)/2.3 x 10^(-9)
≈ 1.44 x 10^(-10) M
Therefore, the concentration of HOBr in the solution is approximately 1.44 x 10^(-10) M.
(c) (i) To calculate the volume of 0.115 M Ba(OH)2 needed to reach the equivalence point when titrated into a 65.0 mL sample of 0.146 M HOBr, we can use the concept of stoichiometry and the balanced chemical equation:
HOBr(aq) + 2 Ba(OH)2(aq) → Ba(OBr)2(aq) + 2 H2O(l)
From the equation, we can see that the mole ratio between HOBr and Ba(OH)2 is 1:2. Therefore, the moles of Ba(OH)2 required can be obtained using the equation:
moles of Ba(OH)2 = (moles of HOBr) x (2 moles of Ba(OH)2/1 mole of HOBr)
moles of HOBr = concentration x volume
= 0.146 M x 0.065 L
= 0.00949 moles
moles of Ba(OH)2 = 0.00949 moles x 2
= 0.01898 moles
Using the molarity (0.115 M) and the moles calculated, we can determine the volume of Ba(OH)2:
volume of Ba(OH)2 = moles of Ba(OH)2 / Molarity
= 0.01898 moles / 0.115 M
= 0.165 L or 165 mL
Therefore, the volume of 0.115 M Ba(OH)2 solution needed to reach the equivalence point is approximately 165 mL.
(ii) At the equivalence point, a strong base (Ba(OH)2) reacts with a weak acid (HOBr) to form a salt (Ba(OBr)2). As a result of this reaction, the pH at the equivalence point will be greater than 7 (basic). This is because the strong base will neutralize the weak acid, resulting in an excess of hydroxide ions (OH-) in solution.