# What is the #"pH"# of a #"0.49-M"# #"NaClO"# solution?

##### 1 Answer

#### Explanation:

Sodium hypochlorite is **soluble** in water, so it dissociates completely in aqueous solution to produce sodium cations,

This means that you have

#["ClO"^(-)]_ 0 = ["NaClO"] = "0.49 M"#

The hypochlorite anions will act as a **weak base** in aqueous solution, meaning that some of these anions will react with water to form hypochlorous acid,

#"HClO"_ ((aq))^(-) + "H"_ 2"O"_ ((l)) rightleftharpoons "HClO"_ ((aq)) + "OH"_ ((aq))^(-)#

So right from the start, you should expect the

Now, the **base dissociation constant** of the hypochlorite anions is calculated using the equation

#K_b = (["HClO"] * ["OH"^(-)])/(["ClO"^(-)])#

So if you start with **ionize** to produce hypochlorous acid and hydroxide anions, then the **at equilibrium**, you have

#["HClO"] = ["OH"^(-)] = x quad "M"# If

#x# #"M"# of hypochlorite anions ionize, then you get#x# #"M"# of hypochlorous acid and#x# #"M"# of hydroxide anions.

#["ClO"^(-)] = (0.49 - x) quad "M"# If

#x# #"M"# of hypochlorite anions ionize, then you are left with#(0.49 - x)# #"M"# of hypochlorite anions.

The expression of the base dissociation constant will be

#K_b = (x * x)/(0.49 - x)#

#K_b = x^2/(0.49 - x)#

Now, you didn't provide a value for the base dissociation constant, but I can tell you that it's *small enough* to allow you to use the approximation

#0.49 - x ~~ 0.49#

This means that you have

#K_b = x^2/0.49#

and so

#x = sqrt(0.49 * K_b)#

Since we've said that **equilibrium concentration** of hydroxide anions, you will have

#["OH"^(-)] = sqrt(0.49 * K_b) quad "M"#

As you know, an aqueous solution at

#"pH + pOH = 14"#

Since

#"pOH" = - log(["OH"^(-)])#

you can say that the

#"pH" = 14 + log(["OH"^(-)])#

In your case, this will be equal to

#"pH" = 14 + log(sqrt(0.49 * K_b))#

Now all you have to do is to use the value of the base dissociation constant given to you--or you can simply do a quick search for the base dissociation constant of the hypochlorite anion--and plug it into the equation to find the

The answer *should* be rounded to two **decimal places** because you have two **sig figs** for the concentration of the solution.

Alternatively, you can look up the **acid dissociation constant**,

#K_a * K_b = 1.0 * 10^(-14) implies K_b = (1.0 * 10^(-14))/K_a#

This will get you

#"pH" = 14 + log(sqrt(0.49 * (1.0 * 10^(-14))/K_a))#