Question #66b60

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Question #66b60

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Answer 1 · 2015-05-15T20:57:02

Here's what's going to happen this time. The balanced chemical equation for your neutralization reaction was

$C H_{3} C O O H_{(a q)} + O H_{(a q)}^{-} \to C H_{3} C O O_{(a q)}^{-} + H_{2} O_{(l)}$ $CH_3COOH_((aq)) + OH_((aq))^(-) -> CH_3COO_((aq))^(-) + H_2O_((l))$

Calculate how many moles of acetic acid and of sodium hydroxide you add together

$C = \frac{n}{V} \Rightarrow n = C \cdot V$ $C = n/V => n = C * V$

$n_{C H_{3} C O O H} = 0.100 M \cdot 25 \cdot 10^{- 3} L = 0.0025 moles$ $n_(CH_3COOH) = "0.100 M" * 25 * 10^(-3)"L" = "0.0025 moles"$

$n_{N a O H} = 0.100 M \cdot 25 \cdot 10^{- 3} L = 0.0025 moles$ $n_(NaOH) = "0.100 M" * 25 * 10^(-3)"L" = "0.0025 moles"$

Since you add equal numbers of moles of acetic acid and sodium hydroxide, both compounds will be completely consumed by the reaction.

In the process, the reaction will produce 0.0025 moles of acetate anions, $C H_{3} C O O^{-}$ $CH_3COO^(-)$ .

$n_{C H_{3} C O O H} = 0.0025 - 0.0025 = 0$ $n_(CH_3COOH) = 0.0025 - 0.0025 = 0$

$n_{N a O H} = 0.0025 - 0.0025 = 0$ $n_(NaOH) = 0.0025 - 0.0025 = 0$

$n_{C H_{3} C O O^{-}}^{-} = 0.0025 moles$ $n_(CH_3COO^(-)) = "0.0025 moles"$

The acetate anion can act as a weak base and react with water to produce hydroxide ions and reform acetic acid according to the following equilibrium

$C H_{3} C O O_{(a q)}^{-} + H_{2} O_{(l)} ⇌ C H_{3} C O O H_{(a q)} + O H_{(a q)}^{-}$ $CH_3COO_((aq))^(-) + H_2O_((l)) rightleftharpoons CH_3COOH_((aq)) + OH_((aq))^(-)$

To determine the concentration of the acetate anion, use the total volume of the solution, which is

$V_{sol} = V_{acetic acid} + V_{sodium hydroxide}$ $V_"sol" = V_"acetic acid" + V_"sodium hydroxide"$

$V_{sol} = 25 + 25 = 50. mL$ $V_"sol" = 25 + 25 = "50. mL"$

This will get you

$[C H_{3} C O O^{-}] = \frac{0.0025 moles}{50 . \cdot 10^{- 3} L} = 0.0500 M$ $[CH_3COO^(-)] = "0.0025 moles"/(50. * 10^(-3)"L") = "0.0500 M"$

Use an ICE table to determine the concentration of the hydroxide ions in solution

$C H_{3} C O O_{(a q)}^{-} + H_{2} O_{(l)} ⇌ C H_{3} C O O H_{(a q)} + O H_{(a q)}^{-}$ $" "CH_3COO_((aq))^(-) + H_2O_((l)) rightleftharpoons CH_3COOH_((aq)) + OH_((aq))^(-)$
I.......0.0500.............................................0.............................0
C........(-x).................................................(+x)..........................(+x)
E......0.0500-x..........................................x..............................x

By definition, the base dissociation constant, $K_{b}$ $K_b$ , will be

$K_{b} = \frac{[O H^{-}] \cdot [C H_{3} C O O H]}{[C H_{3} C O O^{-}]} = \frac{x \cdot x}{0.0500 - x} = \frac{x^{2}}{0.0500 - x}$ $K_b = ([OH^(-)] * [CH_3COOH])/([CH_3COO^(-)]) = (x * x)/(0.0500 - x) = x^2/(0.0500 - x)$

Since $K_{b}$ $K_b$ is so small, you can approximate (0.0500 - x) with 0.0500. This will get you

$K_{b} = \frac{x^{2}}{0.0500} = 5.6 \cdot 10^{- 10} \Rightarrow x = 5.29 \cdot 10^{- 6}$ $K_b = x^2/0.0500 = 5.6 * 10^(-10) => x = 5.29 * 10^(-6)$

Since $x = [O H^{-}]$ $x = [OH^(-)]$ , you have

$[O H^{-}] = 5.29 \cdot 10^{- 6} M$ $[OH^(-)] = 5.29 * 10^(-6)"M"$

The pOH of the solution will be

$p O H = - log ([O H^{-}]) = - log (5.29 \cdot 10^{- 6}) = 5.28$ $pOH = -log([OH^(-)]) = -log(5.29 * 10^(-6)) = 5.28$

Therefore, the pH of the solution will be

$p H_{sol} = 14 - p O H = 14 - 5.28 = 8.72$ $pH_"sol" = 14 - pOH = 14 - 5.28 = color(green)(8.72)$

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