Question

How many moles of sodium hypochlorite should be added to 1L of 0.05M hypochlorous acid to form a buffer solution with pH 9.15ssume no changes of volume occurs when NaOCl salt is added?

Answer 1

`"2.1 moles NaClO"`

Explanation:

First thing first - you need to know the acid dissociation constant for hypochlorous acid, `"HClO"`, which is listed as being equal to

`K_a = 3.0 * 10^(-8)`

http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf

So, look at the information provided to you and try to get a feel for what you're dealing with.

You know that sodium hypochlorite, `"NaClO"`, is the salt of hypochlorous acid's conjugate base, which is the hypochlorite anion, `"ClO"^(-)`.

Your solution will thus contain hypochlorous acid, a weak acid, and its conjugate base, which means that you're dealing with a buffer solution.

Before moving forward, calculate the acid's `pK_a`

`color(blue)(pK_a = - log(K_a))`

In your csae,

`pK_a = - log(3.0 * 10^(-8)) = 7.52`

Notice the the pH of the target solution is higher than the acid's `pK_a`, which means that it must contain more conjugate base than weak acid.

But how much more conjugate base?

To figure that out, you can use the Henderson - Hasselbalch equation, which looks like this

`color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"]))`

You know that your solution has

`["HClO"] = "0.05 M"`

Which means that you have everything you need to find `["ClO"^(-)]` by using the H - H equation.

Plug in your values to get

`9.15 = 7.52 + log( (["ClO"^(-)])/(["HClO"]))`

`1.63 = log( (["ClO"^(-)])/(["HClO"]))`

This is equivalent to

`10^log( (["ClO"^(-)])/(["HClO"])) = 10^1.63`

`(["ClO"^(-)])/(["HClO"]) = 42.66`

Therefore,

`["ClO"^(-)] = 42.66 * ["HClO"]`

`["ClO"^(-)] = 42.66 * "0.05 M" = "2.133 M"`

Since you are told the the volume of the solution does not change upon the addition of the salt, you can use its molarity and volume to determine how many moles of salt must be added.

Keep in mind, sodium hypochlorite dissociates in a `1:1` mole ratio with the hypochlorite anion, so

`["ClO"^(-)] = ["NaClO"]`

`color(blue)(c = n/V implies n = c * V)`

`n_(NaClO) = "2.133 M" * "1 L" = "2.133 moles ClO"^(-)`

You should round this off to one sig fig, but I will leave the answer rounded to two sig figs

`n_(NaClO) = color(green)("2.1 moles")`