The neutralisation reaction is :
sf(CH_3COOH+OH^(-)rarrCH_3COO^(-)+H_2O)
Since sf(0.400"M"color(white)(x)Ba(OH)_2) is sf(0.800"M") with respect to sf(OH^(-) ions we can calculate the initial moles of sf(OH^(-)rArr)
sf(nOH^(-)=cxxv=0.800xx35.0/1000=0.028)
The initial moles of sf(CH_3COOH) is given by:
sf(n_(CH_3COOH_("init"))=cxxv=0.650xx60.0/1000=0.039)
From the equation we can see that the no. moles sf(CH_3COO^(-)) formed must be sf(0.028).
The ethanoic acid is INXS so the number of moles remaining is given by:
sf(n_(CH_3COOH)=0.039-0.028=0.011)
This means we have effectively created a buffer solution.
Ethanoic acid dissociates:
sf(CH_3COOHrightleftharpoonsCH_3COO^(-)+H^(+))
For which:
sf(K_a=([CH_3COO^(-)][H^(+)])/([CH_3COOH])=1.75xx10^(-5)color(white)(x)"mol/l")
These are equilibrium concentrations.
To find the sf(pH) we need to find the equilibrium concentration of sf(H^+).
Rearranging gives:
sf([H^+]=K_axx([CH_3COOH])/([CH_3COO^(-)]))
Now check the value of sf(K_a). If it lies between sf(10^(-5)) and sf(10^(-10)) we can assume that, because the dissociation is small, the initial moles is a good approximation to the equilibrium moles.
I will assume sf(K_a) is close enough within this range.
The other thing we can do to make things easier for ourselves is to use moles instead of concentrations.
This is because the total volume is common to both so cancels.
Putting in the numbers sf(rArr)
sf([H^(+)]=1.75xx10^(-5)xx(0.011)/(0.028)=0.685xx10^(-5)color(white)(x)"mol/l")
sf(pH=-log[H^+]=-log(0.685xx10^(-5))
sf(pH=5.16)
