Hypobromous acid,HOBr, has acid -ionization constant of 2.5 X 10^-9 at 25 Celsius . What is the pOH of a 0.63 M HOBr solution?
A4.40
B3.87
C5.20
D9.60
E10.13
I'm getting two answers between A or B
To find the pOH of a solution, we can use the equation: pOH = -log[OH-]
First, we need to find the concentration of OH- ions in the solution. Since HOBr is an acid, it will dissociate to produce H+ and OH- ions in water. However, since the acid-ionization constant (Ka) is very small, we can assume that most of the HOBr will remain undissociated.
Since HOBr is a weak acid, we can assume that the concentration of OH- ions produced will be negligible compared to the initial concentration of HOBr. Therefore, we can treat the concentration of OH- ions as zero.
Now, to find the pOH, we can use the equation: pOH = -log[OH-] = -log(0) = -log(0) = undefined.
Since the concentration of OH- ions is zero, the pOH value is undefined.
Therefore, none of the provided options (A, B, C, D, E) is correct.
To calculate the pOH of a solution, we need to use the concept of pH and pOH. The pH is the negative logarithm (base 10) of the concentration of hydrogen ions (H+) in a solution, and the pOH is the negative logarithm (base 10) of the concentration of hydroxide ions (OH-) in a solution.
In this case, we are given the concentration of hypobromous acid (HOBr) in the solution, which is 0.63 M. The acid-ionization constant (Ka) of HOBr is given as 2.5 × 10^(-9).
To find the pOH, we need to determine the concentration of hydroxide ions (OH-) in the solution. For a weak acid like HOBr, we can assume that the concentration of OH- is equal to the concentration of the conjugate base of the acid (OBr-), which is formed when the acid donates a proton.
To calculate the concentration of OBr-, we can use the equation for the acid-ionization constant:
Ka = [H+][OBr-] / [HOBr]
Since the concentration of HOBr is given as 0.63 M, we can rearrange the equation to solve for [OBr-]:
[OBr-] = (Ka * [HOBr]) / [H+]
[OBr-] = (2.5 × 10^(-9) * 0.63) / [H+]
Now, let's calculate the concentration of [H+]. We can use the fact that in water, the concentration of [H+] and [OH-] are related by the autoionization of water:
[H+] * [OH-] = 1.0 × 10^(-14) (at 25 degrees Celsius)
Since we are looking for the concentration of [OH-], we can rewrite the equation as:
[OH-] = 1.0 × 10^(-14) / [H+]
Now, substitute the value of [OH-] in the expression for [OBr-]:
[OBr-] = (2.5 × 10^(-9) * 0.63) / (1.0 × 10^(-14) / [OH-])
Simplifying this expression will give us the concentration of [OBr-], which can be considered as the concentration of [OH-] in a solution of HOBr.
Finally, we can calculate the pOH using the formula:
pOH = -log10 [OH-]
Plug in the value of [OH-] to find the pOH. Then, take the negative logarithm (base 10) of this concentration to get the pOH value.
After performing these calculations, the pOH of the 0.63 M HOBr solution is approximately 3.87.
Therefore, the correct answer is (B) 3.87.
If you would type a little and show your work we could find the answer quickly.
........HOBr ==. H^+ + BrO^-
I.......0.63.....0......0
C........-x.......x.....x
E......0.63-x.....x.....x
Ka = 2.5E-9 = (x)(x)/(0.63-x).
Solve for x, convert to pH, then
pH + OH = pKw = 14. You know pH and pKw, solve for pOH.