Question #c3719

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Question #c3719

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Answer 1 · 2015-08-24T14:11:05

$p H_{sol} = 3.37$ $pH_"sol" = 3.37$

Explanation:

So, you start with a solution that contains formic acid, $HCOOH$ $"HCOOH"$ , a weak acid, to which you add sodium hydroxide, $NaOH$ $"NaOH"$ , a strong base.

To get a better understanding of what happens when these two solutions are mixed, start by calculating the pH of the formic acid solution.

Use an ICE table and the molarity of the acid to help you with that

${HCOOH}_{(aq]} + H_{2} O_{(l]} ⇌ {HCOO}_{(aq]}^{-} + H_{3} O_{(aq]}^{+}$ $"HCOOH"_text((aq]) + "H"_2"O"_text((l]) rightleftharpoons "HCOO"_text((aq])^(-) + "H"_3"O"_text((aq])^(+)$

$I 0.100 00$ $color(purple)("I")" " " " 0.100" " " " " " " " " " " " " " " "0" " " " " " " " 0$
$C (- x) (+ x) (+ x)$ $color(purple)("C")" " " "(-x)" " " " " " " " " " " " " "(+x)" " " " "(+x)$
$E (0.100-x) x x$ $color(purple)("E")" " "(0.100-x) " " " " " " " " " " " " " " "x" " " " " " " "x$

By definition, the acid dissociation constant for this equilibrium will be equal to

$K_{a} = \frac{[{HCOO}^{-}] \cdot [H_{3} O^{+}]}{[HCOOH]}$ $K_a = (["HCOO"^(-)] * ["H"_3"O"^(+)])/(["HCOOH"])$

$K_{a} = \frac{x \cdot x}{0.100 - x} = \frac{x^{2}}{0.100 - x}$ $K_a = (x * x)/(0.100 - x) = x^2/(0.100-x)$

You can approximate $(0.100 - x)$ $(0.100-x)$ wit $0.100$ $0.100$ because $K_{a}$ $K_a$ is small enough. This means that you have

$1.8 \cdot 10^{- 4} = \frac{x^{2}}{0.100} \Rightarrow x = \sqrt{0.18 \cdot 1.8 \cdot 10^{- 4}} = 4.24 \cdot 10^{- 3} M$ $1.8 * 10^(-4) = x^2/0.100 implies x = sqrt(0.18 * 1.8 * 10^(-4)) = 4.24 * 10^(-3)"M"$

The pH of the solution will thus be

$p H_{sol} = - log ({[H_{3} O]}_{3}^{+})$ $pH_"sol" = - log(["H"_3"O"]^(+))$

$p H_{sol} = - log (4.24 \cdot 10^{- 3}) = 2.37$ $pH_"sol" = -log(4.24 * 10^(-3)) = 2.37$

Now you add the strong base. The weak acid will react with the strong base to produce sodium formate, $HCOONa$ $"HCOONa"$ , which will exist in aqueous solution as $N a^{+}$ $Na"^(+)$ and ${HCOO}^{-}$ $"HCOO"^(-)$ , and water.

${HCOOH}_{(aq]} + {OH}_{(aq]} \to HCOO_{(aq]} + H_{2} O_{(l]}$ $"HCOOH"_text((aq]) + "OH"_text((aq]) -> "HCOO"""_text((aq]) + "H"_2"O"_text((l])$

Calculate how many moles of each you have

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

$n_{N a O H} = 0.150 M \cdot 5.00 \cdot 10^{- 3} L = 0.00075 moles NaOH$ $n_(NaOH) = "0.150 M" * 5.00 * 10^(-3)"L" = "0.00075 moles NaOH"$

and

$n_{H C O O H} = 0.100 M \cdot 25.00 \cdot 10^{- 3} L = 0.0025 moles HCOOH$ $n_(HCOOH) = "0.100 M" * 25.00 * 10^(-3)"L" = "0.0025 moles HCOOH"$

Since you have more acid than strong base, the strong base will be completely consumed by the reaction.

This means that you'll be left with

$n_{H C O O H} = 0.0025 - 0.00075 = 0.00175 moles HCOOH$ $n_(HCOOH) = 0.0025 - 0.00075 = "0.00175 moles HCOOH"$

The reaction produced

$n_{H C O O^{-}}^{-} = n_{N a O H} = 0.00075 moles HCOO^{-}$ $n_(HCOO^(-)) = n_(NaOH) = "0.00075 moles HCOO"""^(-)$

This means that you've created a buffer, which is a solution that (in your case) contains a weak acid and its conjugate base in comparable amounts.

To make the calculations easier, I'll use the Henderson-Hasselbalch equation to determine the pH of the resulting solution

$p H_{sol} = p K_{a} + log (\frac{[{CHOO}^{-}]}{[HCOOH]})$ $pH_"sol" = pK_a + log( (["CHOO"^(-)])/(["HCOOH"]))$

The concentrations of the weak acid and of its conjugate base in the final solution will be

$V_{final} = V_{H C O O H} + V_{N a O H}$ $V_"final" = V_(HCOOH) + V_(NaOH)$

$V_{final} = 25.00 + 5.00 = 30.0 mL$ $V_"final" = 25.00 + 5.00 = "30.0 mL"$

$[{CHOO}^{-}] = \frac{0.00075 moles}{30.0 \cdot 10^{- 3} L} = 0.0250 M$ $["CHOO"^(-)] = "0.00075 moles"/(30.0 * 10^(-3)"L") = "0.0250 M"$

and

$[HCOOH] = \frac{0.00175 moles}{30.0 \cdot 10^{- 3} L} = 0.0583 M$ $["HCOOH"] = "0.00175 moles"/(30.0 * 10^(-3)"L") = "0.0583 M"$

The acid's $p K_{a}$ $pK_a$ will be

$p K_{a} = - log (K_{a})$ $pK_a = -log(K_a)$

$p K_{a} = - log (1.8 \cdot 10^{- 4}) = 3.74$ $pK_a = - log(1.8 * 10^(-4)) = 3.74$

This means that the solution's pH will be

$pH_"sol" = 3.74 + log((0.0250color(red)(cancel(color(black)("M"))))/(0.0583color(red)(cancel(color(black)("M"))))) = color(green)(3.37)$

I strongly recommend calculating the pH without using the Henderson-Hasselbalch equation by using the ICE table again, only this time taking into account the fact that you have some conjugate base present from the neutralization reaction.

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