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# How many moles of sodium hypobromite (NaBrO) must be added to a 2.00 L solution of 0.025 M hypobromous acid (HBrO) to form a buffer solution of pH 9.0 (Ka HBrO is 2.5x10-9)? Assume the volume of solution does not change on addition of NaBrO?

##"17 g"##

A 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, ##"NaBrO"##, is needed in order to supply enough hypobromite anions, the **conjugate base** of hypobromous acid, ##"HBrO"##, in order to form a buffer of equal to ##9##.

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 of the hypobromite anions must be, then use the volume of the solution to find how many moles it must contain.

The ##pK_a## of hypobromous acid is

##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 ##"0.025 M"##.

Remember, your target pH is ##9##, so you can say that

##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 , the answer will be

##m_"NaBrO" = color(green)("17 g")##