# Question #7f583

##### 1 Answer

Here's what I got.

#### Explanation:

The key to this problem is the balanced chemical equation for this neutralization reaction.

*weak acid*, will react with *sodium hydroxide*, **strong base**, to produce water and the salt of the acid's conjugate base,

Since you're in aqueous solution, you can simplify the equation by removing the sodium cations,

#"HX"_text((aq]) + "NaOH"_text((aq]) -> "X"_text((aq])^(-) + "H"_2"O"_text((aq])#

Notice that you have

This tells you that *for every* one mole of strong base added to the weak acid solution, **one mole** of weak acid is *consumed* and **one mole** of conjugate base is *produced*.

So, your initial solution contains

#1/3 * n -># one thirdof the number of moles of weak acid

In order to consume *solution B* will be equal to

#n_"sol B" = n - n/3 = color(green)((2n)/3)#

Remember that consuming **produce** ** B** will contain

#"For HX: " (2n)/3" moles"#

#"For X"^(-): color(white)(a)n/3" moles"#

At this point, you should be able to tell that the concentration of the weak acid is **twice as high** as the concentration of the conjugate base in solution ** B**. Nevertheless, here's how you can prove that.

Let's assume that the **volume of solution B** is equal to

#color(blue)(c = n/V)#

#["HX"] = (2n)/3"moles" * 1/"V L" = ((2n)/3 * 1/V) " M"#

#["X"^(-)] = n/3"moles" * 1/"V L" = (n/3 * 1/V)" M"#

This means that you have

#(["HX"])/(["X"^(-)]) = ((2n)/3 * color(red)(cancel(color(black)(1/V " M"))))/(n/3 * color(red)(cancel(color(black)(1/V " M")))) = (2 color(red)(cancel(color(black)(n))))/3 * 3/color(red)(cancel(color(black)(n))) = color(green)(2)#

Now, in order to find the *acid dissociation constant*,

#"HX"_text((aq]) + "H"_2"O"_text((aq]) rightleftharpoons "H"_3"O"_text((aq])^(+) + "X"_text((aq])^(-)#

By definition,

#K_a = (["H"_3"O"^(+)] * ["X"^(-)])/(["HX"]) = (["X"^(-)])/(["HX"]) * ["H"_3"O"^(+)]#

Since you know that

#["H"_3"O"^(+)] = 4.2 * 10^(-4)"M"#

and that

#(["HX"])/(["X"^(-)]) = 2 <=> (["X"^(-)])/(["HX"]) = 1/2#

you can say that

#K_a = 1/2 * 4.2 * 10^(-4) = color(green)(2.1 * 10^(-4))#

Finally, *methyl orange* is **not** a suitable indicator for this titration because at the equivalence point,m the pH of the resulting solution is **higher** than

This happens because when all of the weak acid is neutralized, the remaining solution will still contain its conjugate base

Methyl orange changes color in the

which means that it will not be useful for your weak acid - strong base titration. Instead, you should pick an indicator that changes color *above*