An acetate buffer was made by dissolving CH_3CO_2H in water, then using NaOH to

Question

An acetate buffer was made by dissolving CH_3CO_2H in water, then using NaOH to convert some of the acetic acid (weak acid or "WA") to its conjugate base ("CB"). The resulting solution has pH = 4.50 and a total concentration ([WA] + [CB]) of 0.050 M. The K_a for CH_3CO_2H is 1.8 Times 10^-5. Write the net ionic equation for the acid-base reaction used to prepare this buffer. Calculate the ratio of [CB]/[WA] in this buffer. Using the ratio you calculated in part (b) and the total buffer concentration given above, solve for the actual concentrations of WA and CB in this buffer. Calculate the moles of WA and CB in 100. mL of this buffer. How many moles of each would you have after the addition of 0.00050 mol NaOH? Calculate the resulting pH.

Explanation / Answer

CH3COOH + NaOH ------> CH3COONa + H2O

CH3COO- + H+ + Na+ + OH- ------> CH3COO- + Na+ + H2O(l)

net ionic equation H+ + OH- ------->H2O

Ka = 1.8*10-5

Pka = -logka

         = -log1.8*10-5

        = 4.75

PH = Pka + log[CB]/[WA]

4.5 = 4.75+ log[CB]/[WA]

4.5-4.75 = log[CB]/[WA]

-0.25   = log[CB]/[WA]

[CB]/[WA] = 10-0.25 = 0.562

[CB] = 0.562[WA]

[WA]+[CB] = 0.05M

[WA] + 0.562[WA] = 0.05M

1.562[WA] = 0.05

[WA] = 0.05/1.562 = 0.032M

[CB] = 0.562[WA]

[CB] = 0.562*0.032 = 0.0179M

no of moles of WA = molarity * volume in L

                             = 0.032*0.1 = 0.0032 moles

no of moles of CB   =0.0179*0.1 = 0.00179moles

By the addition of 0.0005 moles NaOH

no of moles of WA = 0.0032-0.0005 =0.0027 moles

no of moles of CB = 0.00179+ 0.0005 =0.00229 moles

PH = PKa +log[CB]/[WA]

    = 4.75 + log0.00229/0.0027

   = 4.75-0.07152 =4.6784

                            

      

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