# Question 503ff

Jan 5, 2016

$\text{pH} = p {K}_{a}$

#### Explanation:

So, you're dealing with a buffer solution that contains a weak acid, $\text{HA}$, and its conjugate base, ${\text{A}}^{-}$, delivered to the solution by its salt, $\text{NaA}$.

Before doing any calculation, try to estimate what the pH of the buffer will be relative to the acid's $p {K}_{a}$.

Notice that you have $0.20$ moles of weak acid and $0.20$ moles of salt in solution. Since the salt dissociates in a $1 : 1$ mole ratio with the conjugate base of the acid

${\text{NaA"_text((aq]) -> "Na"_text((aq])^(+) + "A}}_{\textrm{\left(a q\right]}}^{-}$

you can say that the solution contains $0.20$ moles of conjugate base.

Since the volume of the solution is the same for both species, you can say that the solution will have equal concentrations of weak acid and of conjugate.

When this is the case, the pH of the buffer will be equal to the acid's $p {K}_{a}$.

You can prove this by using the Henderson - Hasselbalch equation for a weak acid - conjugate base buffer

color(blue)("pH" = pK_a + log( (["A"^(-)])/(["HA"])))" ", where

$\left[{\text{A}}^{-}\right]$ - the concentration of the conjugate base
$\left[\text{HA}\right]$ - the concentration of the weak acid

As you can see here, when you have equal concentrations of conjugate base and weak acid, the log term will be equal to $0$, since $\log \left(1\right) = 0$, and the pH will be equal to the $p {K}_{a}$.

So right from the start, you can say that the pH of this buffer will be equal to $4.8$.

The actual concentrations of the weak acid and of the conjugate base will be

["HA"] = ["A"^(-)] = "0.20 moles"/"1 dm"^3 = "0.20 M"

Therefore,

"pH" = pK_a + log( color(red)(cancel(color(black)("0.20 M")))/color(red)(cancel(color(black)("0.20 M"))))

$\text{pH} = \textcolor{g r e e n}{4.8}$

Notice that you can use this equation to predict how the pH of the buffer will change when you have

• more weak acid than conjugate base

log( (["A"^(-)])/(["HA"])) < 0 implies "pH" < pK_a

The solution will be more acidic.

• more conjugate base than weak acid

log( (["A"^(-)])/(["HA"])) > 0 implies "ph" > pK_a

The solution will be more basic.

Jan 5, 2016

pH = 4.8

#### Explanation:

Write out the acid dissociation equation:
HA + H_2O ⇌ A^(-) + H_3O^+#

You have 0.2 mol acid in $1 {\mathrm{dm}}^{3}$, therefore [HA] = 0.2 mol/L.
You have 0.2 mol salt in $1 {\mathrm{dm}}^{3}$, therefore [NaA] = 0.2 mol/L, AND the A- ion will also have a concentration of 0.2 mol/L, because this salt completely dissociates into its ions.

Now we put this information into an ICE chart:

You can eliminate the x in the "0.2-x" and "0.2+x" expressions, because the Ka value in weak acids is very small.

Write out the Ka expression:
${K}_{a} = \frac{\left[{A}^{-}\right] \left[{H}_{3} {O}^{+}\right]}{\left[H A\right]}$
${K}_{a} = \frac{\left(0.2\right) x}{0.2}$
${K}_{a} = x$

Since you need to find the pH, you have to use $- \log \left[{H}_{3} {O}^{+}\right]$ or $- \log \left[x\right]$

Therefore:
$p H = - \log \left[x\right]$
$p H = - \log \left[{K}_{a}\right]$
$p H = p K a$
$p H = 4.8$