# Question #545aa

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

The idea here is that neutralizing a *weak acid* with a **strong base** can produce a * buffer solution, depending on whether or not the neutralization is complete*.

A weak acid will react with a strong base to produce water and the **conjugate base** of the acid.

If the neutralization is **complete**, the resulting solution will contain the conjugate base of the acid, which will then react with water to reform some of the weak acid and produce *hydronium ions*,

If the neutralization **is not complete**, then the resulting solution will contain a weak acid and its conjugate base **buffer solution**.

The pH of a buffer solution that contains a weak acid and its conjugate base can be calculated using the **Henderson - Hasselbalch equation**

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

Here

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

So, your problem didn't provide you with concentrations of weak acid and strong base because you don't *actually* need to know them in order to find the pH of the resulting solution given that a **certain percent of the weak acid** is neutralized.

Let's assume that your weak acid is

#"HA"_text((aq]) + "OH"_text((aq])^(-) -> "A"_text((aq])^(-) + "H"_2"O"_text((l])#

It's important to notice that you have **mole ratios** between the weak acid, the strong base, and the conjugate base

This means that **for every** one mole of weak acid neutralized by one mole of strong base, **one mole** of conjugate base is produced.

Now, let's assume that you want to neutralize **after** the reaction is finished.

If you start with a volume

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

#n_"before" = ["HA"]_0 * V_1#

After the reaction is complete, you need to have

#n_"after" = 6/10 * ["HA"]_0 * V_1 -># moles of weak acid remaining in solution

In order to neutralize

#n_"base" = 4/10 * ["HA"]_0 * V_1 -># moles of strong base needed for the reaction

Now, if **produce**

Let's say that the volume of strong base is **total volume** of the solution will be

#V_"total" = V_1 + V_2#

The concentration of the weak acid will now be

#["HA"] = overbrace(6/10 * ["HA"]_0 * V_1)^(color(purple)("moles of weak acid remaining in solution")) * 1/(V_1 + V_2)#

The concentration of the conjugate base will be

#["A"^(-)] = overbrace(4/10 * ["HA"]_0 * V_1)^(color(green)("moles of conjugate base produced")) * 1/(V_1 + V_2)#

Now, plug these values into the Henderson - Hasselbalch equation to get

#"pH" = pK_a + log((["A"^(-)])/(["HA"]))#

#"pH" = pK_a + log( (4/10 * color(red)(cancel(color(black)(["HA"]_0 * V_1))))/(6/10 * color(red)(cancel(color(black)(["HA"]_0 * V_1)))) * color(red)(cancel(color(black)(V_1 + V_2)))/color(red)(cancel(color(black)(V_1 + V_2))))#

#"pH" = pK_a + log(4/color(red)(cancel(color(black)(10))) * color(red)(cancel(color(black)(10)))/6)#

#"pH" = pK_a + log(2/3)#

*Now, does this result make sense?*

If you neutralize **more weak acid** than conjugate base. This means that the pH of the solution will be **lower** than the

Since

#"pH" = pK_a -0.18#

you have will indeed have

#"pH" < pK_a#

As practice, try redoing the calculations if you **higher** than