5.00×10−3mol of HBr are dissolved in water to make 15.0L of solution. What is the concentration of hydroxide ions, [OH−], in this solution?
HBr is a strong acid; i.e., it is ionized 100%. Therefore, (H^+) = 5E-3 mols/15.0L = ?M
Then (H^+)(OH^-) = Kw = 1E-14.
You know Kw and you know (H^+), solve for (OH^-)
To determine the concentration of hydroxide ions ([OH-]) in this solution, we need to consider the dissociation of HBr in water.
HBr dissociates into H+ (hydrogen ion) and Br- (bromide ion). Since we are interested in the concentration of hydroxide ions, we need to determine the concentration of H+ ions first.
Given that the initial amount of HBr dissolved is 5.00×10^(-3) mol and the final volume of the solution is 15.0 L, we can calculate the initial concentration of HBr:
Initial concentration of HBr = moles/volume
= 5.00×10^(-3) mol / 15.0 L
= 3.33×10^(-4) mol/L
Since HBr is a strong acid, it completely dissociates to release 1 H+ ion for every HBr molecule. Therefore, the concentration of H+ ions is the same as the initial concentration of HBr:
[H+] = 3.33×10^(-4) mol/L
Now, we need to consider the autoprotolysis of water. In pure water, some water molecules act as bases and accept protons (H+) to form hydroxide ions (OH-). The concentration of H+ ions in pure water is 1×10^(-7) mol/L.
However, the H+ concentration in our solution is higher due to the presence of HBr. To calculate the concentration of hydroxide ions ([OH-]) in the solution, we need to use the equation for the autoprotolysis of water:
[H+][OH-] = 1.0×10^(-14) mol^2/L^2
So, we can rearrange the equation as:
[OH-] = 1.0×10^(-14) mol^2/L^2 / [H+]
Now, substitute the value of [H+] into the equation:
[OH-] = 1.0×10^(-14) mol^2/L^2 / (3.33×10^(-4) mol/L)
Calculating this will give us the concentration of hydroxide ions ([OH-]) in the solution.