# Question 46acb

Feb 8, 2015

So, you start with $\text{200 mL}$ of $\text{0.15 M}$ potassium acetate ($C {H}_{3} C O O K$) solution. In order to determine the pH of this solution you must use the given acid dissociation constant for acetic acid, since you're dealing with one of its salts.

Use the ICE chart method (more here: http://en.wikipedia.org/wiki/RICE_chart) to determine the concentration of ${\text{OH}}^{-}$ in the solution

$C {H}_{3} C O {O}_{\left(a q\right)}^{-} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s C {H}_{3} C O O {H}_{\left(a q\right)} + O {H}_{\left(a q\right)}^{-}$
I.....0.15 M...................................................0...........................0
C......(-x)......................................................(+x).......................(+x)
E.....0.15-x....................................................x...........................x

SInce $C {H}_{3} C O O K$ is the conjugate base of $C {H}_{3} C O O H$, you must use the base dissociation constant, or ${K}_{b}$, which is equal to

${K}_{b} = {K}_{w} / {K}_{a} = {10}^{- 14} / \left(1.8 \cdot {10}^{- 5}\right) = 5.6 \cdot {10}^{- 10}$

Therefore, ${K}_{b} = \frac{\left[C {H}_{3} C O O H\right] \cdot \left[O {H}^{-}\right]}{\left[C {H}_{3} C O {O}^{-}\right]} = \frac{x \cdot x}{0.15 - x}$

SIDE NOTE. I won't go through the calculations because I'll try to make the answer as short as possbile.

The value of $x$ will be $9.2 \cdot {10}^{- 6} \text{M}$, which means that the solution's pH will be

$p {H}_{1} = 14 - \left(- \log \left(\left[O {H}^{-}\right]\right)\right) = 14 - 5.04 = 8.96$

Now you start adding the other solutions one by one. $\text{KOH}$ will fully dissociate in aqueous solution, which means that the concentration of $O {H}^{-}$ added will be equal to the concentration of $K O H$ added.

$K O {H}_{\left(a q\right)} r i g h t \le f t h a r p \infty n s {K}_{\left(a q\right)}^{+} + O {H}_{\left(a q\right)}^{-}$

Determine the molarity of the solution by using its $\text{m/m}$ percent concentration.

$\text{5.6%} = {m}_{s o l u t e} / \left({m}_{s o l u t i o n}\right) \cdot 100$

Since you've got $\text{45 mL}$ of a solution that has a density of $\text{1.045 g/mL}$, the weight of the sample will be

$m = \rho \cdot V = \text{1.045 g/mL" * "45 mL" = "47.03 g}$

This means that the mass of $\text{KOH}$ you have will be

${m}_{s o l u t e} = \frac{5.6 \cdot {m}_{s o l u t i o n}}{100} = \frac{5.6 \cdot 47.03}{100} = \text{2.63 g}$

The number of moles of $\text{KOH}$ you have is

$\text{2.63 g" * ("1 mole")/("56.1 g") = "0.0470 moles}$

The concentration of the $\text{KOH}$ solution after you add it will be

C = n/V_("total") = ("0.0470 moles")/((200 + 45)*10^(-3)"L") = "0.19 M"

The total concentration of ${\text{OH}}^{-}$ will now be

$\left[O {H}^{-}\right] = 9.2 \cdot {10}^{- 6} + 0.19 \cong \text{0.19 M}$

The new solution's pH will now be

$p {H}_{2} = 14 - \left(- \log \left(\left[O {H}^{-}\right]\right)\right) = 14 - 0.72 = 13.28$

Now you add the third and (hopefully) final solution. This time you're adding a weak acid to a strong base. The number of moles of acetic acid you'll add to the solution and its concentration will be

$n = C \cdot V = \text{0.855 M" * 55 * 10^(-3)"L" = "0.047 moles}$

C_(acetic) = n/V_("total") = "0.047 moles"/((200 + 45 + 55) * 10^(-3)"L") = "0.16 M"

Notice that the number of moles of $C {H}_{3} C O O H$ and $O {H}^{-}$ are equal, which means that both of these species will be consumed in the reaction and $\text{0.047 moles}$ of $C {H}_{3} C O {O}^{-}$ will be produced. The number of moles of $C {H}_{3} C O {O}^{-}$ that you already have in solution is

$n = C \cdot V = \text{0.15 M" * 300 * 10^(-3)"L" = "0.045 moles}$

Once again, use the ICE table

$C {H}_{3} C O O {H}_{\left(a q\right)} + O {H}_{\left(a q\right)}^{-} \to C {H}_{3} C O {O}_{\left(a q\right)}^{-} + H 2 {O}_{\left(l\right)}$
I.....0.047 moles....0.047 moles.......0.045 moles
C...(-0.047)..............(-0.047)..............+(0.047)
E....0 moles.............0 moles............0.092 moles

This means that the first equilibrium will be reestablished with a new concentration of $C {H}_{3} C O {O}^{-}$

C_("new") = "0.092 moles"/(300 * 10^(-3)"L") = "0.306 M"#

Therefore, the concentration of ${\left[O H\right]}^{-}$ will be

${K}_{b} = \frac{\left[C {H}_{3} C O O H\right] \cdot \left[O {H}^{-}\right]}{\left[C {H}_{3} C O {O}^{-}\right]} = \frac{x \cdot x}{0.306 - x} = 5.6 \cdot {10}^{- 10}$

$\left[O {H}^{-}\right] = 1.39 \cdot {10}^{- 5} \text{M}$

$p {H}_{F I N A L} = 14 - \left(- \log \left(1.39 \cdot {10}^{- 5}\right)\right) = 14 - 4.86 = 9.14$