# 47.25 mL of 0.1120 M Acetic acid titrated with 0.4750 M KOH. Calculate the initial pH, and the pH at the equivalence point of titration? Acetic Acid Dissociation Constant: 1.75 x 10 -5 ​

Feb 23, 2015

The initial pH is $\text{2.85}$ and the equivalence point pH is $\text{8.87}$.

So, you start with a solution of acetic acid. The pH of this solution can be determined by using the concentration of protons in solution. For that, you'll have to set up an ICE chart

....$C {H}_{3} C O O {H}_{\left(a q\right)} r i g h t \le f t h a r p \infty n s C {H}_{3} C O {O}_{\left(a q\right)}^{-} + {H}_{\left(a q\right)}^{+}$
I.....0.1120..................................0.........................0
C....(-x).....................................(+x)......................(+x)
E...(0.1120-x)............................x...........................x

Use the expression of the acid dissociation constant to solve for $x$

${K}_{a} = \frac{x \cdot x}{0.1120 - x} = 1.75 \cdot {10}^{- 5}$

The value you'll get for $x$ is $\text{0.0014}$. Plug this into the pH equation and you'll get

$p {H}_{\text{initial}} = - \log \left(0.0014\right) = 2.85$

Now you begin the titration by adding $K O H$, which is a strong base. As a result, it will react with acetic acid, which is a weak acid, to form water and potassium acetate, a salt of acetic acid, according to this balanced equation

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

Notice that you have a $\text{1:1}$ mole ratio between acetic acid and potassium hydroxide; this means that, at the equivalence point, the number of moles of potassium hydroxide used will be equal to the number of moles of acetic acid already in solution.

The number of moles of acetic acid you have in solution is

n_("acetic") = C * V = "0.1120M" * 47.32*10^(-3)"L" = "0.00530 moles"

Automatically, both the acetic acid and the potassium hydroxide will be consumed in the reaction; the $\text{1:1}$ mole ratio between either of these compounds and potassium acetate implies that you'll produce the same amount of $C {H}_{3} C O O K$ moles as you just consumed.

Now, you'll need to determine the total volume of the solution; for this you must determine how much $K O H$ was added

C = n/V => V_("KOH") = n/C = "0.00530 moles KOH"/"0.4750 M" = "11.16 mL"

The total volume of the solution will be

V_("total") = V_("acetic") + V_("KOH") = "47.25 mL" + "11.16 mL" = "58.41 mL"

Use this volume to determine the concentration of the potassium acetate formed in solution

C = n/V_("total") = "0.00530 moles"/(58.41 * 10^(-3)"L") = "0.09074 M"

After the acetic acid and ptoassium hydroxide are consumed, $C {H}_{3} C O O K$ will react with water to form acetic acid and hydroxide anions; use the ICE chart again to solve for the concentration of hydroxide ions

.....$C {H}_{3} C O {O}_{\left(a q\right)}^{-} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s C {H}_{3} C O O {H}_{\left(a q\right)} + O {H}_{\left(a q\right)}^{-}$
I......0.09704..................................................0.......................0
C.....(-x)..........................................................(+x)...................(+x)
E...(0.09704 - x)............................................x.........................x

This time use the base dissociation constant, or ${K}_{b}$, which equals ${K}_{b} = {K}_{w} / {K}_{a} = {10}^{- 14} / \left(1.75 \cdot {10}^{- 5}\right) = 5.71 \cdot {10}^{- 10}$

Use this to determine $x$

$5.71 \cdot {10}^{- 10} = {x}^{2} / \left(0.09704 - x\right) \implies x = 0.000007444$

Plug this baby into this equation to solve for the final pH

$p {H}_{\text{equiv}} = 14 - p O H = 14 - \log \left(0.000007444\right)$

$p {H}_{\text{equiv}} = 14 - 5.13 = 8.87$