# Question f3e75

Oct 14, 2016

$\textsf{p H = 5.16}$

#### Explanation:

The neutralisation reaction is :

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

Since $\textsf{0.400 \text{M} \textcolor{w h i t e}{x} B a {\left(O H\right)}_{2}}$ is $\textsf{0.800 \text{M}}$ with respect to sf(OH^(-) ions we can calculate the initial moles of $\textsf{O {H}^{-} \Rightarrow}$

$\textsf{n O {H}^{-} = c \times v = 0.800 \times \frac{35.0}{1000} = 0.028}$

The initial moles of $\textsf{C {H}_{3} C O O H}$ is given by:

$\textsf{{n}_{C {H}_{3} C O O {H}_{\text{init}}} = c \times v = 0.650 \times \frac{60.0}{1000} = 0.039}$

From the equation we can see that the no. moles $\textsf{C {H}_{3} C O {O}^{-}}$ formed must be $\textsf{0.028}$.

The ethanoic acid is INXS so the number of moles remaining is given by:

$\textsf{{n}_{C {H}_{3} C O O H} = 0.039 - 0.028 = 0.011}$

This means we have effectively created a buffer solution.

Ethanoic acid 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:

$\textsf{{K}_{a} = \frac{\left[C {H}_{3} C O {O}^{-}\right] \left[{H}^{+}\right]}{\left[C {H}_{3} C O O H\right]} = 1.75 \times {10}^{- 5} \textcolor{w h i t e}{x} \text{mol/l}}$

These are equilibrium concentrations.

To find the $\textsf{p H}$ we need to find the equilibrium concentration of $\textsf{{H}^{+}}$.

Rearranging gives:

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

Now check the value of $\textsf{{K}_{a}}$. If it lies between $\textsf{{10}^{- 5}}$ and $\textsf{{10}^{- 10}}$ we can assume that, because the dissociation is small, the initial moles is a good approximation to the equilibrium moles.

I will assume $\textsf{{K}_{a}}$ is close enough within this range.

The other thing we can do to make things easier for ourselves is to use moles instead of concentrations.

This is because the total volume is common to both so cancels.

Putting in the numbers $\textsf{\Rightarrow}$

$\textsf{\left[{H}^{+}\right] = 1.75 \times {10}^{- 5} \times \frac{0.011}{0.028} = 0.685 \times {10}^{- 5} \textcolor{w h i t e}{x} \text{mol/l}}$

sf(pH=-log[H^+]=-log(0.685xx10^(-5))#

$\textsf{p H = 5.16}$