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Question

A buffer solution is 0.20 M in acetic acid and in sodium acetate. Calculate the change in pH upon adding 1.0 mL of 0.10 M hydrochloric acid to 10.0 mL of this solution. The Ka of acetic acid is 1.75*10^-5

Answer 1

For this problem we need to first understand that you are going to be using the Henderson-Hasselbalch equation for the pH of buffer:  pH = pKa + log( [ A^-] / [HA]) 

 

But in order to use this formula we need to first know how much of the week acid and its complementary base we have in the solution after the addition of HCl. For this we will have to use a stoichiometric ice table to calculate the amount of compounds we have after the addition and reaction between the HCl and the sodium acetate ( weak base)

 

Use Dilution equation to figure the initial change in concentration be the increase in volume.

 

M1V1=M2V2

 

0.10 M HCl ( 1.0 mL) = M2(11.0 mL)

 

*note: "e^number" = x 10^number

 

M2= 9.091e-3 M HCl

 

Now lets do this with the acetic acid and sodium acetate with have the same initial concentration so the initial change should also be the same.

 

0.20 M ( 10.0 mL) = M2 ( 11.0 mL)

 

M2= 0.1818 M acetic acid and sodium acetate

 

 

 

Reaction:   HCl(aq)       +   NaC2H3O2(aq) <--> HC2H3O2(aq) + NaCl(aq)

Before   :  9.091e-3M          0.1818M                        0M                   0M

Change  :-9.091e-3M          -9.091e-3M              +9.091e-3M    +9.091e-3M  

After     :     0 M                 0.1727 M                   9.091e-3M       9.091e-3M  

 

With this we have calculated the final concentration of the sodium acetate and the addition concentration acetic acid

 

[Sodium acetate] = 0.1727 M  

[Acetic Acid] = 9.091e-3M + 0.1818M  = 0.1909M

pKa = - log(Ka) = - log(1.75e-5) = 4.757

 

Finally we can use the final formula

 

pH = pKa + log( [ A-] / [HA])

 

pH= 4.757 + log ([0.1727 M/0.1909M])

 

Plug that into you calculator and you should be fine.