The pH of a 0.55-M aqueous solution of hypobromous acid, HBrO, at 25°C is 4.48. What is the value of Ka for HBrO?
how do you solve for Ka?
2.0X 10^ -9
To solve for Ka, you need to use the equilibrium constant expression for the dissociation of the acid. For hypobromous acid (HBrO), it dissociates according to the following equation:
HBrO(aq) ⇌ H+(aq) + BrO-(aq)
The Ka expression for this reaction is:
Ka = [H+(aq)][BrO-(aq)] / [HBrO(aq)]
In this case, you know the pH of the solution, which allows you to calculate the concentration of H+ ions. Since pH is defined as the negative logarithm of the H+ ion concentration, you can use this relationship to find the concentration of H+ ions:
[H+(aq)] = 10^(-pH)
From the given information, the pH of the solution is 4.48. Substituting this value into the equation, you can solve for [H+(aq)]:
[H+(aq)] = 10^(-4.48)
Next, you need to determine the concentration of HBrO(aq). In this case, the concentration of HBrO is given as 0.55 M.
Finally, using the equilibrium constant expression, you can rearrange the equation to solve for Ka:
Ka = [H+(aq)][BrO-(aq)] / [HBrO(aq)]
Substitute the known values into the equation:
Ka = (10^(-4.48)) * (x) / (0.55)
Here, x represents the concentration of BrO-(aq).
By solving this equation, you can find the value of Ka for HBrO.
To solve for Ka (acid dissociation constant), we need to use the equation for the dissociation of the acid in water, which is as follows:
HBrO(aq) ↔ H+(aq) + BrO-(aq)
The Ka expression for this reaction is:
Ka = [H+(aq)][BrO-(aq)] / [HBrO(aq)]
To determine the value of Ka, we need the concentrations of H+(aq), BrO-(aq), and HBrO(aq). However, we only have the pH of the solution, which is given as 4.48.
To convert the pH to the concentration of H+(aq), we use the equation:
pH = -log[H+(aq)]
Therefore, [H+(aq)] = 10^(-pH)
Now, we need to find the concentration of HBrO(aq). The given solution is 0.55 M, so the concentration of HBrO(aq) is 0.55 M.
Since HBrO(aq) dissociates into H+(aq) and BrO-(aq) in a 1:1 ratio, the concentration of BrO-(aq) is also 0.55 M.
Now that we know the concentrations of [H+(aq)], [BrO-(aq)], and [HBrO(aq)], we can substitute them into the Ka expression to solve for Ka:
Ka = (0.55 M)(0.55 M) / (10^(-4.48))
Now, calculate the value of Ka using these values.
HOBr ==> H^+ + OBr^-
Ka = (H^+)(OBr^-)/(HOBr)
Convert pH to H^+ and substitute into Ka expression. OBr^- is the same as H^+, substitute into Ka expression. HOBr = (0.55-H^+). Substitute into Ka expression. Solve for Ka.