# Question fe671

Oct 24, 2015

$\text{17 g}$

#### 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, $\text{NaBrO}$, is needed in order to supply enough hypobromite anions, the conjugate base of hypobromous acid, $\text{HBrO}$, in order to form a buffer of pH 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 molarity of the hypobromite anions must be, then use the volume of the solution to find how many moles it must contain.

The $p {K}_{a}$ of hypobromous acid is

$p {K}_{a} = - \log \left({K}_{a}\right)$

$p {K}_{a} = - \log \left(2.5 \cdot {10}^{- 9}\right) = 8.60$

Since the aicd dissociation constant is so small, you can assume that the concentration of the acid will be approximately equal to $\text{0.025 M}$.

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

$9.0 = 8.60 + \log \left(\left(\left[\text{BrO"^(-)])/(["HBrO}\right]\right)\right)$

$\log \left(\left(\left[\text{BrO"^(-)])/(["HBrO}\right]\right)\right) = 0.40$

This is equivalent to

$\left(\left[\text{BrO"^(-)])/(["HBrO}\right]\right) = {10}^{0.40} = 2.512$

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

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{\text{NaBrO" = "0.0628 M" * "2.00 L" = "0.1256 moles NaBrO}}$

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")#