# What are the initial pH and the pH at the equivalence if 0.2000 M acetic acid is titrated with 30.41 mL of 0.2000 M NaOH solution? Ka = 1.75 *10(-5)

Feb 18, 2015

So, you start with a $\text{0.20000-M}$ solution of acetic acid, or $C {H}_{3} C O O H$. The initial pH of the solution will depend on the concentration of protons in solution. So, using the ICE table method (more here: http://en.wikipedia.org/wiki/RICE_chart), you get

$C {H}_{3} C O O {H}_{\left(a q\right)} r i g h t \le f t h a r p \infty n s C {H}_{3} C O {O}_{\left(a q\right)}^{-} + {H}_{\left(a q\right)}^{+}$
I.......0.2.................................0........................0
C.....(-x)................................(+x)......................(+x)
E...(0.2 - x)............................x..........................x

${K}_{a} = \frac{\left[{H}^{+}\right] \cdot \left[C {H}_{3} C O {O}^{-}\right]}{\left[C {H}_{3} C O O H\right]} = \frac{x \cdot x}{0.2 - x} = 1.75 \cdot {10}^{- 5}$

Solving for $x$ will give you $x = 0.00187$, which is equal to the concentration of protons in solution. Therefore,

$p {H}_{\text{initial}} = - \log \left(\left[{H}^{+}\right]\right) = - \log \left(0.00187\right) = 2.73$

As you perform the titration, you observe the equivalence point occurs when $\text{30.41 mL}$ of $\text{0.2000-M}$ $N a O H$ solution is added.

Since $N a O H$ is a strong base, it will react with acetic acid, which is a weak acid, to produce sodium acetate, the salt of its conjugate base, and water.

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

Notice the $\text{1:1}$ mole ratio you have between acetic acid and $N a O H$. This means that at the equivalence point, the number of moles of acetic acid will be equal to the number of moles of $N a O H$. So,

${n}_{N a O H} = C \cdot V = \text{0.2000 M" * 30.41 * 10^(-3)"L}$
${n}_{N a O H} = \text{0.00610 moles}$

Automatically,

$\text{0.00610 moles NaOH" * ("1 mole acetic acid")/("1 mole NaOH") = "0.00610 moles acetic acid}$

Since both the initial amount of acetic acid, and the added $N a O H$ will be consumed at the equivalence point, the number of moles of $C {H}_{3} C O O N a$ produced will be equal to $\text{0.00610}$ (mole ratio again $\to$ 1 mole of each reactant, 1 mole of each product).

The initial volume of acetic acid you had is

$C = \frac{n}{V} \implies V = \frac{n}{C} = \text{0.00610 moles"/("0.2000 M") = "30.5 mL}$

This means that the concentration of $C {H}_{3} C O O N a$, which is equivalent to $C {H}_{3} C O {O}^{-}$, will be

C_(CH_3COO^(-)) = n/V_("total") = "0.00610 moles"/((30.5 + 30.41)*10^(-3)"L") = "0.1 M"

$C {H}_{3} C O O N a$ will hydrolize to produce acetic acid and hydroxide, or $O {H}^{-}$. Use the ICE table method again

$C {H}_{3} C O {O}_{\left(a q\right)}^{-} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s C {H}_{3} C O O {H}_{\left(a q\right)} + O {H}_{\left(a q\right)}^{-}$
I.......0.1....................................................0........................0
C.....(-x)..................................................(+x)......................(+x)
E...(0.1 - x)...............................................x..........................x

This time use ${K}_{b}$, which is equal to ${K}_{W} / {K}_{a} = \frac{1.0 \cdot {10}^{- 14}}{1.75 \cdot {10}^{- 5}}$, or
${K}_{b} = 5.71 \cdot {10}^{- 10}$

${K}_{b} = \frac{\left[O {H}^{-}\right] \cdot \left[C {H}_{3} C O O H\right]}{\left[C {H}_{3} C O {O}^{-}\right]} = {x}^{2} / \left(0.1 - x\right)$

Solving for $x$ will get you

$x = \text{0.00000756} = \left[O {H}^{-}\right] = 7.56 \cdot {10}^{- 6}$. Therefore,

$p O H = - \log \left(\left[O {H}^{-}\right]\right) = - \log \left(7.56 \cdot {10}^{- 6}\right) = 5.12$

Now use the fact that $p {H}_{s o l u t i o n} = 14 - p O H$ to get

$p {H}_{\text{final}} = 14 - 5.12 = 8.88$