Part A
6.00×10−3 mol of HBr are dissolved in water to make 16.0L of solution. What is the concentration of hydroxide ions,[OH-], in this solution?
Express your answer with the appropriate units.
To find the concentration of hydroxide ions, [OH-], in the solution, we need to determine the number of hydroxide ions present.
We can start by finding the concentration of HBr using the given information:
Number of moles of HBr = 6.00 × 10^-3 mol
Volume of solution = 16.0 L
The concentration of HBr is calculated using the formula:
Concentration (M) = Number of moles / Volume
Concentration of HBr = 6.00 × 10^-3 mol / 16.0 L
Next, we need to know the stoichiometry of the reaction between HBr and water. HBr is a strong acid, and it completely dissociates in water to form H+ and Br- ions. Since there is no direct information given about the reaction with hydroxide ions (OH-), we need to use the fact that water undergoes autoionization to produce both H+ and OH- ions:
H2O ⇌ H+ + OH-
In pure water, the concentration of H+ ions is equal to the concentration of OH- ions, which is 1.00 × 10^-7 M at 25°C.
However, because we have added HBr to the water, it will increase the concentration of H+ ions (as Br- ions will not affect the OH- concentration). Assuming all the HBr dissociates completely, the concentration of H+ ions will be equal to the concentration of HBr.
Therefore, the concentration of H+ ions (and HBr) is:
Concentration of H+ (and HBr) = 6.00 × 10^-3 mol / 16.0 L
Now, to find the concentration of OH- ions, we can use the relationship between [H+] and [OH-] in water:
Kw = [H+][OH-] = 1.00 × 10^-14
Rearranging the equation, we have:
[OH-] = Kw / [H+]
Substituting the values:
[OH-] = 1.00 × 10^-14 / (6.00 × 10^-3 mol / 16.0 L)
Calculating this expression will give us the concentration of hydroxide ions, [OH-], in the solution.
To find the concentration of hydroxide ions, [OH-], in the solution, we need to first find the number of moles of hydroxide ions present.
The chemical equation for the ionization of HBr in water is:
HBr + H2O -> H3O+ + Br-
From the equation, we can see that one molecule of HBr forms one hydronium ion (H3O+) and one bromide ion (Br-).
Now, let's determine the number of moles of hydroxide ions in the solution.
Since HBr doesn't directly produce hydroxide ions (OH-), we need to find the concentration of hydronium ions (H3O+) and convert it to hydroxide ions.
In a neutral solution, the concentration of hydronium ions is equal to the concentration of hydroxide ions. Therefore, we need to find the concentration of hydronium ions first.
From the given information, we have:
Number of moles of HBr = 6.00 × 10^-3 mol
Volume of the solution = 16.0 L
Concentration of HBr = moles/volume
Concentration of HBr = (6.00 × 10^-3 mol) / (16.0 L)
Concentration of HBr = 3.75 × 10^-4 mol/L
Since the concentration of HBr is the same as the concentration of H3O+ ions in a neutral solution, we have:
Concentration of H3O+ = 3.75 × 10^-4 mol/L
Now, to find the concentration of hydroxide ions, we use the fact that in a neutral solution, the product of the concentrations of hydronium ions and hydroxide ions is equal to the ion product of water (1.0 × 10^-14 at 25°C):
[H3O+][OH-] = 1.0 × 10^-14
Since the solution is neutral, the concentration of hydronium ions is equal to the concentration of hydroxide ions, so we can substitute:
(3.75 × 10^-4 mol/L)([OH-]) = 1.0 × 10^-14
Rearranging the equation to solve for [OH-]:
[OH-] = (1.0 × 10^-14) / (3.75 × 10^-4 mol/L)
[OH-] ≈ 2.67 × 10^-11 mol/L
Therefore, the concentration of hydroxide ions, [OH-], in the solution is approximately 2.67 × 10^-11 mol/L.
M = mols/L = 6.00E-3/16.0 = ? = (H^+)
pH = -log (H^+)
Then pH + pOH = pKw = 14
You know pH and pKw, solve for pOH and from that use pOH = -log(OH^-) and solve for OH^-