# 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?

##### 1 Answer

#### Answer:

#### Explanation:

First thing first - you need to know the *acid dissociation constant* for hypochlorous acid,

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

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*,

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

#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 **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

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

#["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")#