# Calculate the pH of the buffer solution prepared by mixing 10.0 mL of a 0.100M NaOH solution with 30.0 mL of a 0.100M CH3COOH solution. Ka(CH3COOH) = 1.76*10^-5 ?

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36
Aug 9, 2015

Alternatively, you could use the Henderson-Hasselbalch equation.

#### Explanation:

Since you're dealing with a buffer, which in your case is a solution that contains a weak acid, acetic acid, and its conjugate base, the acetate anion, in comparable amounts, you can use the Henderson-Hasselbalch equation to determine its pH.

pH_"sol" = pK_a + log ( (["conjugate base"])/(["weak acid"]))

Here $p {K}_{a}$ is equal to

$p {K}_{a} = - \log \left({K}_{a}\right)$

$p {K}_{a} = - \log \left(1.76 \cdot {10}^{- 5}\right) = 4.75$

Use the classic approach to determine the concentrations of the weak acid and of the conjugate base after the sodium hydroxide solution is added.

Since you add together $1 \cdot {10}^{- 3} \text{moles}$ of sodium hydroxide and $3 \cdot {10}^{- 3} \text{moles}$ of acetic acid, you will be left with

${n}_{\text{acetic acid" = 3 * 10^(-3) - 1 * 10^(-3) = 2 * 10^(-3)"moles}}$

The reaction will produce the same number of moles of acetate ions as the number of moles of sodium hydroxide consumed. This means that you will get

[CH_3COOH] = (2 * 10^(-3)"moles")/(40 * 10^(-3)"L") = "0.05 M"

[CH_3COO^(-)] = (1 * `10^(-3)"moles")/(40 * 10^(-3)"L") = "0.025 M"

Therefore, the solution's pH is

pH_"sol" = 4.75 + log ((0.025cancel("M"))/(0.05cancel("M")))

$p {H}_{\text{sol}} = 4.75 + \left(- 0.301\right) = \textcolor{g r e e n}{4.45}$

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Aug 9, 2015

The pH of the solution is $4.45$.

#### Explanation:

You have ${\text{10 cm}}^{3}$ of a $\text{0.100-moldm-3}$ NaOH solution, which you mix with ${\text{30 cm}}^{3}$ of a $\text{0.100-moldm-3}$ $C {H}_{3} C O O H$ solution.

You know that acid dissociation constant of acetic acid is ${K}_{a} = 1.76 \cdot {10}^{- 5}$.

Work out the number of moles of $N a O H$ first

$\text{0.100 moldm-3" * 10 * 10^(-3)"dm3" = 1 * 10^(-3) "moles}$

then work out the number of moles of $C {H}_{3} C O {O}_{H}$

$\text{0.100 moldm-3" * 30 * 10^(-3)"dm3" = 3 * 10^(-3) "moles}$

Notice that you have less moles of sodium hydroxide than you do of ethanoic acid. This means that all the moles of sodium hydroxide will react with the ethanoic acid to form the salt sodium ethanoate, $C {H}_{3} C O O N a$.

Even more importantly, you will have some ethanoic acid still left in solution. This indicates that you can expect the final solution's pH to be acidic, i.e. smaller than $7$.

The amount in excess $C {H}_{3} C O O H$ is-

$3 \cdot {10}^{- 3} - 1 \cdot {10}^{- 3} = 2 \cdot {10}^{- 3} \text{moles}$

The balanced chemical equation for this reaction looks like this

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

The number of moles of sodium ethanoate produced by the reaction will be equal to the number of moles of sodium hydroxide that reacted.

Since sodium ethanoate actually exists as sodium cations and ethanoate anions in solution, you can say that the number of moles of $C {H}_{3} C O {O}^{-}$ is the same as the number of moles of $N a O H$.

Total volume of the solution is

${V}_{\text{total" = 10 + 30 = "40 cm"^3 = 40 * 10^(-3)"dm}}^{3}$

Ethanoic acid is a weak acid, which means that the following equilibrium is established in aqueous solution

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

By definition, ${K}_{a}$ is

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

This means that you have

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

The concentration of the remaining ethanoic acid and formed ethanoate ions will be

[CH_3COOH] = (2 * 10^(-3)"moles")/(40 * 10^(-3)"dm3") = "0.05 moldm-3"

[CH_3COO^(-)] = (1 * 10^(-3)"M")/(40 * 10^(-3)"dm3") = "0.025 moldm-3"

Therefore, you have

[H^(+)] = 1.76 * 10^(-5) * ("0.05 moldm-3")/"0.025 moldm-3"

$\left[{H}^{+}\right] = 3.52 \cdot {10}^{- 5} \text{moldm-3}$

The pH of the solution will thus be

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

$p {H}_{\text{sol}} = - \log \left(3.62 \cdot {10}^{- 5}\right) = \textcolor{g r e e n}{4.45}$

I have a YouTube channel that may help you more on buffers just visit

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