# You need to prepare an acetate buffer of pH 6.32 from a 0.800 M acetic acid solution and a 2.46 M KOH solution. If you have 925 mL of the acetic acid solution, how many milliliters of the KOH do you need to add to make a bufer of pH 6.32?

## The pKa of acetic acid is 4.76.

Aug 13, 2017

293 ml

#### Explanation:

Ethanoic acid is a weak acid and dissociates:

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

For which:

$\textsf{{K}_{a} = \frac{\left[C {H}_{3} C O {O}^{-}\right] \left[{H}^{+}\right]}{\left[C {H}_{3} C O O H\right]}}$

These are equilibrium concentrations.

$\textsf{p {K}_{a} = 4.76}$ from which $\textsf{{K}_{a} = 1.737 \times {10}^{- 5}}$

Rearranging we get:

$\textsf{\left[{H}^{+}\right] = {K}_{a} \times \frac{\left[C {H}_{3} C O O H\right]}{\left[C {H}_{3} C O {O}^{-}\right]}}$

$\textsf{p H = 6.32}$ from which $\textsf{\left[{H}^{+}\right] = 4.786 \times {10}^{- 7} \textcolor{w h i t e}{x} \text{mol/l}}$

$\therefore$sf(4.786xx10^(-7)=1.737xx10^(-5)xx([CH_3COOH])/([CH_3COO^(-)])

$\therefore$$\textsf{\frac{\left[C {H}_{3} C O O H\right]}{\left[C {H}_{3} C O {O}^{-}\right]} = \frac{4.786 \times {10}^{- 7}}{1.737 \times {10}^{- 5}} = 0.027553}$

To get the required pH we need to set the acid : salt ratio to this value.

To produce the required ethanoate ions we add $\textsf{O {H}^{-}}$ ions:

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

You can see from the equation that the no. moles $\textsf{O {H}^{-}}$ added = the no. of moles of $\textsf{C {H}_{3} C O {O}^{-}}$ produced.

Since we know $\textsf{\left[O {H}^{-}\right]}$ we can say that if $\textsf{V}$ is the volume of the $\textsf{O {H}^{-}}$ solution then:

$\textsf{n O {H}^{-} = n C {H}_{3} C O {O}_{\text{eqm}}^{-} = 2.26 \times V}$

Where $\textsf{n}$ is the no. moles.

We know the initial moles of ethanoic acid:

$\textsf{n C {H}_{3} C O O {H}_{\text{init}} = 0.800 \times 0.925 = 0.740}$

So we can get the moles after the alkali has been added. Because the dissociation is small we will assume that these are a good approximation to the equilibrium moles:

$\textsf{n C {H}_{3} C O O {H}_{\text{eqm}} = \left(0.740 - 2.46 V\right)}$

$\therefore$$\textsf{\left(n C {H}_{3} C O O {H}_{\text{eqm"))/(nCH_3COO_("eqm}}^{-}\right) = \frac{0.740 - 2.46 V}{2.46 V} = 0.027553}$

We can use moles and not concentrations since the total volume is common to both so cancels anyway.

$\therefore$$\textsf{0.740 - 2.46 V = 0.027553 \times 2.46 V}$

$\textsf{0.06778 V + 2.46 V = 0.740}$

$\textsf{V = \frac{0.740}{2.5277} = 0.293 \textcolor{w h i t e}{x} L}$

$\textsf{V = 293 \textcolor{w h i t e}{x} \text{ml}}$

$- - - - - - - - - - - - - - - - - - - -$

Check by iteration:

Volume KOH = 297 ml

$\textsf{n O {H}^{-} = 0.293 \times 2.46 = 0.72078}$

$\therefore$$\textsf{n C {H}_{3} C O {O}_{\text{eqm}}^{-} = 0.72078}$

$\textsf{n C {H}_{3} C O O {H}_{\text{init}} = 0.740}$

$\therefore$$\textsf{n C {H}_{3} C O O {H}_{\text{eqm}} = 0.740 - 0.72078 = 0.01922}$

$\therefore$ sf("ratio"" "( ["acid"] )/ (["salt"]) = 0.01922/0.72078 = 0.0266655)

$\textsf{\left[{H}^{+}\right] = 1.737 \times {10}^{- 5} \times 0.0266655 = 0.046318 \times {10}^{- 5} \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{p H = 6.33}$