Question #109c2
1 Answer
Explanation:
Start by calculating the
"p"K_a = - log(K_a)pKa=−log(Ka)
You will have
"p"K_a = - log(2.3 * 10^(-9)) = 8.64pKa=−log(2.3⋅10−9)=8.64
Now, notice that the pH of the buffer is slightly higher than the
In other words, you should expect to find
(["BrO"^(-)])/(["HBrO"]) > 1[BrO−][HBrO]>1
The pH of a buffer that contains a weak acid and its conjugate base can be calculated by using the Henderson - Hasselbalch equation
color(blue)(ul(color(black)("pH" = - "p"K_a + log( (["conjugate base"])/(["weak acid"])))))
For your buffer, you have
"pH" = "p"K_a + log((["BrO"^(-)])/(["HBrO"]))
which is
8.75 = 8.64 + log((["BrO"^(-)])/(["HBrO"]))
Rearrange the equation as
log((["BrO"^(-)])/(["HBrO"])) = 8.75 - 8.64
This will be equivalent to
10^log((["BrO"^(-)])/(["HBrO"])) = 10^(0.11)
which gets you
color(darkgreen)(ul(color(black)((["BrO"^(-)])/(["HBrO"]) = 1.3)))
The answer is rounded to two sig figs, the number of decimal places you have for the pH of the solution.
As predicted, the buffer contains more conjugate base than weak acid.
