# Question #fe671

##### 1 Answer

#### 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*, *hypobromite anions*, the **conjugate base** of *hypobromous acid*,

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"]))#

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)#

#pK_a = - 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"]))#

#log( (["BrO"^(-)])/(["HBrO"])) = 0.40#

This is equivalent to

#(["BrO"^(-)])/(["HBrO"]) = 10^0.40 = 2.512#

Therefore, you found that

#["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 * V#

#n_"NaBrO" = "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")#