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

The pKa of acetic acid is 4.76.

1 Answer
Aug 13, 2017

Answer:

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.