Question #87990

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Question #87990

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Answer 1 · 2015-05-14T21:31:23

Here's what I got.

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So, you're titrating hypochlorous acid, a weak acid, with potassium hydroxide, a strong base. The pH of the solution at those specific points in the titration will be

Before adding any $K O H$ $KOH$

Right from the get-go, since you're dealing with a weak acid, you can expect the pH of the solution to be acidic.

The hypochlorous acid will react with water molecules to form hydronium ions, $H_{3} O^{+}$ $"H"_3"O"^(+)$ , and hypochlorite ions, ${ClO}^{-}$ $"ClO"^(-)$ , establishing an equilibrium in the process. Use an ICE table to determine the concentration of the hydronium ions in solution

$H C l O_{(a q)} + H_{2} O_{(l)} ⇌ H_{3} O_{(a q)}^{+} + C l O_{(a q)}^{-}$ $HClO_((aq)) + H_2O_((l)) " "rightleftharpoons" " H_3O_((aq))^(+) " "+" " ClO_((aq))^(-)$

$I 0.150 00$ $color(purple)("I")" " 0.150" " " " " " " " " " " " " " " " " " " " "0" " " " " " " " " " " " " "0$
$C (- x) (+ x) (+ x)$ $color(purple)("C")" "(-x)" " " " " " " " " " " " " " " " " "(+x)" " " " " " " "(+x)$
$E 0.150 - x x x$ $color(purple)("E")" "0.150-x" " " " " " " " " " " " " " " "x" " " " " " " " " " " "x$

By definition, the acid dissociation constant, $K_{a}$ $K_a$ , will be equal to

$K_{a} = \frac{[H_{3} O^{+}] \cdot [{ClO}^{-}]}{[HClO]} = \frac{x \cdot x}{0.150 - x} = \frac{x^{2}}{0.150 - x}$ $K_a = (["H"_3"O"^(+)] * ["ClO"^(-)])/(["HClO"]) = (x * x)/(0.150 - x) = x^2/(0.150 - x)$

Because $K_{a}$ $K_a$ is so small, you can approximate (0.150 - x) with 0.150. This means that you'll get

$K_{a} = \frac{x^{2}}{0.150} = 4.0 \cdot 10^{- 8} \Rightarrow x = 7.75 \cdot 10^{- 5}$ $K_a = x^2/0.150 = 4.0 * 10^(-8) => x = 7.75 * 10^(-5)$

This will be equal to the concentration of hydronium ions

$[H_{3} O^{+}] = 7.75 \cdot 10^{- 5} M$ $["H"_3"O"^(+)] = 7.75 * 10^(-5)"M"$

The pH of the solution will be

$p H_{sol} = - log ([H_{3} O^{+}]) = - log (7.75 \cdot 10^{- 5}) = 4.11$ $pH_"sol" = -log(["H"_3"O"^(+)]) = -log(7.75 * 10^(-5)) = color(green)(4.11)$

After you add 25 mL of $K O H$ $KOH$

Potassium hydroxide will react with the hypochlorous acid to produce hypochlorite ions. In the process, some of the weak acid will be consumed, along with the added strong base.

$H C l O_{(a q)} + K O H_{(a q)} \to K C l O_{(a q)} + H_{2} O_{(l)}$ $HClO_((aq)) + KOH_((aq)) -> KClO_((aq)) + H_2O_((l))$

Calculate the number of moles of $HClO$ $"HClO"$ you initially had in solution by using its molarity and volume

$C = \frac{n}{V} \Rightarrow n = C \cdot V$ $color(blue)(C = n/V => n = C *V)$

$n_{H C l O} = 50.0 \cdot 10^{- 3} L \cdot 0.150 M = 0.0075 moles HClO$ $n_(HClO) = 50.0 * 10^(-3)"L" * "0.150 M" = "0.0075 moles HClO"$

The number of moles of potassium hydroxide will be

$n_{K O H} = C \cdot V = 25 \cdot 10^{- 3} L \cdot 0.150 M = 0.00375 moles KOH$ $n_(KOH) = C * V = 25 * 10^(-3)"L" * "0.150 M" = "0.00375 moles KOH"$

According to the balanced chemical equation, 1 mole of $KOH$ $"KOH"$ will consume 1 mole of $HClO$ $"HClO"$ and produce 1 mole of ${ClO}^{-}$ $"ClO"^(-)$ .

This means that the solution will now contain

$n_{H C l O} = 0.0075 - 0.00375 = 0.00375 moles HClO$ $n_(HClO) = 0.0075 - 0.00375 = "0.00375 moles HClO"$

$n_{K O H} = 0$ $n_(KOH) = 0$

$n_{C l O^{-}}^{-} = {0.00375 moles ClO}^{-}$ $n_(ClO^(-)) = "0.00375 moles ClO"^(-)$

This is the buffer region, i.e. at this point in the titration, your solution contains a weak acid, $HClO$ $"HClO"$ , and its conjugate base, ${ClO}^{-}$ $"ClO"^(-)$ .

This means that you can use the Henderson-Hasselbalch equation to solve for the pH of the solution. Determine the concentration of the species present first

$[HClO] = \frac{n}{V_{total}} = \frac{0.00375 moles}{(50.0 + 25) \cdot 10^{- 3} L} = 0.0500 M$ $["HClO"] = n/V_"total" = "0.00375 moles"/((50.0 + 25) * 10^(-3)"L") = "0.0500 M"$

$[{ClO}^{-}] = [HClO] = 0.0500 M$ $["ClO"^(-)] = ["HClO"] = "0.0500 M"$

The $p K_{a}$ $pK_a$ will be equal to

$p K_{a} = - log (K_{a}) = - log (4.0 \cdot 10^{- 8}) = 7.4$ $pK_a = -log(K_a) = -log(4.0 * 10^(-8)) = 7.4$

Therefore,

$p H_{sol} = p K_{a} + log (\frac{[C l O^{-}]}{[H C l O]})$ $pH_"sol" = pK_a + log(([ClO^(-)])/([HClO]))$

$pH_"sol" = 7.4 + log((cancel("0.0500 M"))/(cancel("0.0500 M"))) = color(green)(7.4)$

When $"pH" = pK_a$ , you've reached the half-equivalence point of the titration.

After adding 40 mL of $KOH$

This is equivalent to adding an extra 15 mL of strong base. The number of moles of base added will be

$n_(KOH) = C * V = "0.150 M" * 15 * 10^(-3)"L" = "0.00225 moles KOH"$

The number of moles of $"HClO"$ left in solution will be

$n_(HClO) = 0.00375 - 0.00225 = "0.0015 moles HClO"$

At the same time,

$n_(KOH) = 0$

$n_(ClO^(-)) = 0.00375 + 0.00225 = "0.00600 moles ClO"^(-)$

The concentrations of the two species will be

$["HClO"] = "0.0015"/((75 + 15) * 10^(-3)"L") = "0.0167 M"$

$["ClO"^(-)] = "0.00600 moles"/((75 + 15) * 10^(-3)"L") = "0.0667 M"$

The pH of the solution will be

$pH_"sol" = 7.4 + log((0.0667cancel("M"))/(0.0167cancel("M"))) = color(green)(8.00)$

After adding 50 mL of $KOH$

This is equivalent to adding another 10 mL of $"KOH"$ . Once again,

$n_(KOH) = 10.0 * 10^(-3)"L" * "0.150 M" = "0.0015 moles KOH"$

This time, all of the weak acid will be consumed and you'll be left with its conjugate base, $"ClO"^(-)$ .

$n_(HClO) = 0.0015 - 0.0015 = 0$

$n_(KOH) = 0$

$n_(ClO^(-)) = 0.00600 + 0.0015 = "0.0075 moles ClO"^(-)$

The concentration of the hypochlorite ions will be

$[ClO^(-)] = "0.0075 moles"/((90 + 10) * 10^(-3)"L") = "0.0750 M"$

This time, the hypochlorite ion will react with water to reform some of the hypochlorous acid, $"HClO"$ , and produce hydroxide ions, $"OH"^(-)$ , according to the following equilibrium reaction

$ClO_((aq))^(-) + H_2O_((l)) " "rightleftharpoons" " OH_((aq))^(-) " "+" " HClO_((aq))$

$color(purple)("I")" " 0.0759" " " " " " " " " " " " " " " " " " "0" " " " " " " " " " " " " "0$
$color(purple)("C")" "(-x)" " " " " " " " " " " " " " " "(+x)" " " " " " " "(+x)$
$color(purple)("E")color(white)(a)0.0759-x" " " " " " " " " " " " " " " "x" " " " " " " " " " " "x$

This time, use the base dissociation constant, $K_b$ , to determine the concentration of the hydroxide ions.

$K_b = K_W/K_a = 10^(-14)/(4.0 * 10^(-8)) = 2.5 * 10^(-7)$

$K_b = ([OH^(-)] * [HClO])/([ClO^(-)]) = x^2/(0.0750) => x = 1.37 * 10^(-4)$

The pOH of the solution will be

$pOH = -log([OH^(-)]) = -log(1.37 * 10^(-4)) = 3.86$

As a result, the pH of the solution will be

$pH_"sol" = 14 - pOH = 14 - 3.86 = color(green)(10.14)$

This will be the pH of the solution at equivalence point. Notice that the complete neutralization of the acid leaves you with its conjugate base present in solution.

Since the solution now contains a weak base, you can expect the pH to be greater than $7$ , which is what happens here.

After adding 60 mL of $KOH$

Once again, this is equivalent to adding another 10 mL of strong base. Once again, the strong base will react with the newly formed weak acid.

The number of moles of hypochlorous acid present in solution will be

$n = C * V = 1.37 * 10^(-4)"M" * 100 * 10^(-3)"L" = "0.0000137 moles"$

The number of moles of $"KOH"$ you add will once again be equal to 0.0015. This means that you'll get

$n_(HClO) = 0$

$n_(OH^(-)) = 0.0015 - 0.0000137 = "0.001486 moles OH"^(-)$

The concentration of hydroxide ions will thus be

$["OH"^(-)] = "0.001486 moles"/((100 + 10) * 10^(-3)"L") = "0.0135 M"$

The pOH will be

$pOH = -log(0.0135) = 1.87$

As a result, the pH of the final solution will be

$pH_"sol" = 14 - pOH = 14 - 1.87 = color(green)(12.13)$

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