# 200 ml of 0.1 M sodium hydroxide is mixed with 100 ml of 0.2 M ethanoic acid. What is the pH of the resulting solution?

Dec 2, 2016

$\textsf{p H = 8.8}$

#### Explanation:

The alkali neutralises the acid:

$\textsf{C {H}_{3} C O O H + N a O H \rightarrow C {H}_{3} C O {O}^{-} N {a}^{+} + {H}_{2} O}$

The moles reacting (n) of each species is given by:

$\textsf{{n}_{O {H}^{-}} = c \times v = 0.1 \times \frac{200}{1000} = 0.02}$

$\textsf{{n}_{C {H}_{3} C O O H} = c \times v = 0.2 \times \frac{100}{1000} = 0.02}$

This shows that they are added in the same mole ratio as defined by the equation.

We can, therefore, say that the no. moles of $\textsf{C {H}_{3} C O {O}^{-}}$ formed = 0.02.

A salt formed from a weak acid and a strong base is slightly alkaline due to hydrolysis:

$\textsf{C {H}_{3} C O {O}^{-} + {H}_{2} O r i g h t \le f t h a r p \infty n s C {H}_{3} C O O H + O {H}^{-}}$

From an ICE table we can use this expression which applies to a weak base:

sf(pOH=1/2(pK_b-logb)

Where $\textsf{b}$ is the concentration of the base which, in this case, is $\textsf{C {H}_{3} C O {O}^{-}}$.

The total volume is now 200 ml + 100 ml = 300 ml = 0.3 L

$\textsf{\left[C {H}_{3} C O {O}^{-}\right] = \frac{n}{v} = \frac{0.02}{0.3} = 0.0666 \textcolor{w h i t e}{x} \text{mol/l}}$

You need to look up the $\textsf{p {K}_{a}}$ value for ethanoic acid which = 4.76.

To get $\textsf{p {K}_{b}}$ you use:

$\textsf{p {K}_{a} + p {K}_{b} = 14}$

$\therefore$$\textsf{p {K}_{b} = 14 - p {K}_{a} = 14 - 4.76 = 9.24}$

Putting in the numbers:

sf(pOH=1/2[9.24-log(0.0666)]

$\textsf{p O H = \frac{1}{2} \left[9.24 - \left(- 1.176\right)\right] = 5.208}$

$\textsf{p H + p O H = 14}$

$\therefore$$\textsf{p H = 14 - p O H = 14 - 5.208 = 8.8}$

As expected the solution is slightly alkaline, even though the acid and base have been added in the same molar ratio.

Dec 2, 2016

$\text{pH} = 8.8$

#### Explanation:

This is really a two-part question:

1. What is in the mixture after the reaction is complete?
2. What is the $\text{pH}$ of the mixture?

What's in the mixture?

I like to use an ICE table to keep track of the stoichiometry calculations.

$\textcolor{w h i t e}{m m m m l l} \text{HAc" +color(white)(l) "OH" → "H"_2"O" + "Ac"^"-}$; K_text(a) = 1.76 × 10^"-5"
$\text{I/mol:} \textcolor{w h i t e}{m l} 0.020 \textcolor{w h i t e}{m l} 0.020 \textcolor{w h i t e}{m m m m m l l} 0$
$\text{Cmol:"color(white)(m)"-0.020"color(white)(m)"-0.020"color(white)(mmmm)"+0.020}$
$\text{E/mol:} \textcolor{w h i t e}{m m} 0 \textcolor{w h i t e}{m m m l} 0 \textcolor{w h i t e}{m m m m m l l} 0.020$

$\text{Initial moles HAc" = 0.100 color(red)(cancel(color(black)("L"))) × "0.2 mol"/(1 color(red)(cancel(color(black)("L")))) = "0.020 mol}$

$\text{Initial moles OH"^"-"= 0.200 color(red)(cancel(color(black)("L"))) × "0.1 mol"/(1 color(red)(cancel(color(black)("L")))) = "0.020 mol}$

We have 300 mL of a solution that contains 0.020 mol of $\text{Ac"^"-}$.

2. What's the $\text{pH}$ of the solution?

$\text{HAc}$ is a weak acid, and $\text{Ac"^"-}$ is its conjugate base.

The solution will be basic.

${K}_{\text{b" = K_"w"/K_"a" = (1.00 × 10^"-14")/(1.76 × 10^"-5") = 5.68 × 10^"-10}}$

Now, we set up another ICE table (with molarities) to calculate the $\text{pH}$ at equilibrium.

["Ac"^"-"] = "0.020 mol"/"0.300 L" = "0.067 mol/L"

$\textcolor{w h i t e}{X X X X X X} \text{Ac"^"-" + "H"_2"O" ⇌ "HAc" + "HO"^"-}$
$\text{I/mol:} \textcolor{w h i t e}{X m} 0.067 \textcolor{w h i t e}{X X X X X X l l} 0 \textcolor{w h i t e}{m m l l} 0$
$\text{C/mol:"color(white)(mml) "-"xcolor(white)(XXXXXXlll) "+"xcolor(white)(mll)"+} x$
$\text{E/mol:"color(white)(Xl)"0.067-} x \textcolor{w h i t e}{X X X X X l l l} x \textcolor{w h i t e}{m m l l} x$

K_b = (["HAc"]["HO"^"-"])/(["Ac"^"-"]) = x^2/(0.067-x) = 5.68 × 10^"-10"

Check if x ≪ 0.067.

0.067/(2.10 × 10^"-10") = 3.2 × 10^8 > 400; ∴ we can ignore $x$.

x^2/0.067 = 5.68 × 10^"-10"

x^2 = 0.067 × 5.68 × 10^"-10" = 3.8 × 10^"-11"

x = 6.2 × 10^"-6"

["HO"^"-"] = 6.2 × 10^"-6" color(white)(l)"mol/L"

"pOH" = "-log"(6.2 × 10^"-6") = 5.21

$\text{pH" = "p"K_"w" - "pOH" = "14.00 - 5.21} = 8.8$