# If 30 mL of 0.10 M NaOH is added to 40 mL of 0.20 M HC2H3O2, what is the pH of the resulting solution at 25°C? Ka for HC2H3O2 is 1.8 x 10^–5 at 25°C.

Apr 24, 2018

See below:

#### Explanation:

The reaction that will occur is:

$N a O H \left(a q\right) + C {H}_{3} C O O H \left(a q\right) \to C {H}_{3} C O O N a + {H}_{2} O \left(l\right)$

Now, using the concentration formula we can find the amount of moles of $N a O H$ and Acetic acid:

$c = \frac{n}{v}$

For $N a O H$
Remember that $v$ should be in litres, so divide any milliliter values by 1000.

$c v = n$

$0.1 \times 0.03 = 0.003 m o l$ of $N a O H$

For $C {H}_{3} C O O H$:

$c v = n$

$0.2 \times 0.04 = 0.008 m o l$ of $C {H}_{3} C O O H$.

So 0.003 mol of $N a O H$ will react to completion with the acid to form 0.003 mol of Sodium acetate, $C {H}_{3} C O O N a$, in the solution, along with 0.005 mol of acid dissolved in a total volume of 70 ml. This will create an acidic buffer solution.

Let's find the concentration of the salt and the acid, respectively:

${c}_{a c i d} = \frac{0.005}{0.7} \approx 0.0714 m o l {\mathrm{dm}}^{-} 3$

${c}_{s a l t} = \frac{0.003}{0.007} \approx 0.0428 m o l {\mathrm{dm}}^{-} 3$

Now, we can use theHenderson-Hasselbalch equation to find the $p H$ of the resulting solution.

The equation looks like this:

$p H = p K a + {\log}_{10} \left(\frac{\left[S a l t\right]}{\left[A c i d\right]}\right)$

We are given the ${K}_{a}$ of the acid, so the $p K a$ is the negative logarithm of the ${K}_{a}$ value.

$p K a = - {\log}_{10} \left[{K}_{a}\right]$
$p K a = - {\log}_{10} \left[1.8 \times {10}^{-} 5\right]$
$p K a \approx 4.74$

Now we have to plug in all the values into the equation:

$p H = 4.74 + {\log}_{10} \left(\frac{\left[0.0428\right]}{\left[0.0714\right]}\right)$

$p H = 4.74 - 0.2218$

$p H \approx 4.52$