Question #fe671
1 Answer
Explanation:
A buffer solution must contain either a Weak acid and its conjugte base, or a weak base and its conjugate acid, in comparable amounts.
The idea here is that you need to determine how much sodium hypobromite,
Your tool of choice here will be the Henderson - Hasselbalch equation for a weak acid - conjugate base buffer
color(blue)("pH" = "pK"_a + log( (["conjugate base"])/(["weak acid"]))pH=pKa+log([conjugate base][weak acid])
You will use this equation to determine what the molarity of the hypobromite anions must be, then use the volume of the solution to find how many moles it must contain.
The
pK_a = - log(K_a)pKa=−log(Ka)
pK_a = - log(2.5 * 10^(-9)) = 8.60pKa=−log(2.5⋅10−9)=8.60
Since the aicd dissociation constant is so small, you can assume that the concentration of the acid will be approximately equal to
Remember, your target pH is
9.0 = 8.60 + log( (["BrO"^(-)])/(["HBrO"]))9.0=8.60+log([BrO−][HBrO])
log( (["BrO"^(-)])/(["HBrO"])) = 0.40log([BrO−][HBrO])=0.40
This is equivalent to
(["BrO"^(-)])/(["HBrO"]) = 10^0.40 = 2.512[BrO−][HBrO]=100.40=2.512
Therefore, you found that
["BrO"^(-)] = ["HBrO"] * 2.512 = "0.025 M" * 2.512 = "0.0628 M"[BrO−]=[HBrO]⋅2.512=0.025 M⋅2.512=0.0628 M
If you assume that the volume of the solution did not change after the addition of the sodium hypobromite, you can say that
C = n/V implies n = C * VC=nV⇒n=C⋅V
n_"NaBrO" = "0.0628 M" * "2.00 L" = "0.1256 moles NaBrO"nNaBrO=0.0628 M⋅2.00 L=0.1256 moles NaBrO
Finally, use sodium hypobromite's molar mass to find how many grams would contain this many moles
0.1256color(red)(cancel(color(black)("moles"))) * "135.9 g"/(1color(red)(cancel(color(black)("mole")))) = "17.069 g"
Rounded to two sig figs, the answer will be
m_"NaBrO" = color(green)("17 g")
