You need to prepare an acetate buffer of pH 6.32 from a 0.800 M acetic acid solution and a 2.46 M $K O H$ $KOH$ solution. If you have 925 mL of the acetic acid solution, how many milliliters of the $K O H$ $KOH$ do you need to add to make a bufer of pH 6.32?

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You need to prepare an acetate buffer of pH 6.32 from a 0.800 M acetic acid solution and a 2.46 M $K O H$ $KOH$ solution. If you have 925 mL of the acetic acid solution, how many milliliters of the $K O H$ $KOH$ do you need to add to make a bufer of pH 6.32?

The pKa of acetic acid is 4.76.

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Answer 1 · 2017-08-13T17:32:51

293 ml

Explanation:

Ethanoic acid is a weak acid and dissociates:

$sf(CH_3COOHrightleftharpoonsCH_3COO^(-)+H^+)$

For which:

$sf(K_a=([CH_3COO^(-)][H^+])/([CH_3COOH]))$

These are equilibrium concentrations.

$sf(pK_a=4.76)$ from which $sf(K_a=1.737xx10^(-5))$

Rearranging we get:

$sf([H^+]=K_axx([CH_3COOH])/([CH_3COO^-]))$

$sf(pH=6.32)$ from which $sf([H^+]=4.786xx10^(-7)color(white)(x)"mol/l")$

$:.$ $sf(4.786xx10^(-7)=1.737xx10^(-5)xx([CH_3COOH])/([CH_3COO^(-)])$

$:.$ $sf(([CH_3COOH])/([CH_3COO^(-)])=(4.786xx10^(-7))/(1.737xx10^(-5))=0.027553)$

To get the required pH we need to set the acid : salt ratio to this value.

To produce the required ethanoate ions we add $sf(OH^-)$ ions:

$sf(CH_3COOH+OH^(-)rarrCH_3COO^(-)+H_2O)$

You can see from the equation that the no. moles $sf(OH^-)$ added = the no. of moles of $sf(CH_3COO^(-))$ produced.

Since we know $sf([OH^-])$ we can say that if $sf(V)$ is the volume of the $sf(OH^-)$ solution then:

$sf(nOH^(-)=nCH_3COO_("eqm")^(-)=2.26xxV)$

Where $sf(n)$ is the no. moles.

We know the initial moles of ethanoic acid:

$sf(nCH_3COOH_("init")=0.800xx0.925=0.740)$

So we can get the moles after the alkali has been added. Because the dissociation is small we will assume that these are a good approximation to the equilibrium moles:

$sf(nCH_3COOH_("eqm")=(0.740-2.46V))$

$:.$ $sf((nCH_3COOH_("eqm"))/(nCH_3COO_("eqm")^(-))=(0.740-2.46V)/(2.46V)=0.027553)$

We can use moles and not concentrations since the total volume is common to both so cancels anyway.

$:.$ $sf(0.740-2.46V=0.027553xx2.46V)$

$sf(0.06778V+2.46V=0.740)$

$sf(V=0.740/2.5277=0.293color(white)(x)L)$

$sf(V=293color(white)(x)"ml")$

$--------------------$

Check by iteration:

Volume KOH = 297 ml

$sf(nOH^(-)=0.293xx2.46=0.72078)$

$:.$ $sf(nCH_3COO_("eqm")^(-)=0.72078)$

$sf(nCH_3COOH_("init")=0.740)$

$:.$ $sf(nCH_3COOH_("eqm")=0.740-0.72078=0.01922)$

$:.$ $sf("ratio"" "( ["acid"] )/ (["salt"]) = 0.01922/0.72078 = 0.0266655)$

$sf([H^+]=1.737xx10^(-5)xx0.0266655=0.046318xx10^(-5)color(white)(x)"mol/l")$

$sf(pH=6.33)$

So not too bad.

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You need to prepare an acetate buffer of pH 6.32 from a 0.800 M acetic acid solution and a 2.46 M $K O H$ $KOH$ solution. If you have 925 mL of the acetic acid solution, how many milliliters of the $K O H$ $KOH$ do you need to add to make a bufer of pH 6.32?

The pKa of acetic acid is 4.76.

1 Answer

Explanation:

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Science

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Humanities

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You need to prepare an acetate buffer of pH 6.32 from a 0.800 M acetic acid solution and a 2.46 M KOHKOHKOH solution. If you have 925 mL of the acetic acid solution, how many milliliters of the KOHKOHKOH do you need to add to make a bufer of pH 6.32?

The pKa of acetic acid is 4.76.

1 Answer

Explanation:

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Impact of this question

You need to prepare an acetate buffer of pH 6.32 from a 0.800 M acetic acid solution and a 2.46 M $K O H$ $KOH$ solution. If you have 925 mL of the acetic acid solution, how many milliliters of the $K O H$ $KOH$ do you need to add to make a bufer of pH 6.32?