Question

a) Beginning with the fully unprotonated lysine, how many mL of 12.00 M HCI will need to be added to create 750.00 ml of a 100.00 mM lysine buffer, pH 10.25?

b) What would the new pH be if 1.00 mL of 5.00 M NaOH were added to the above buffer? What would be the pH if the 1.00 mL of 5.00 M NaOH were added to the 750.00 mL of water?

Answer 1

lysine + 2HCl ----> salt

pH of lysine buffer = pka2 + log(lysine/salt)

pka2 = 8.95

total no of mol of buffer = 750*100*10^-3 = 75 mmol

No of mol of lysine left in buffer = 75-2x mmol

No of mol of HCl must add = salt = 2x mmol

10.25 = 8.95+log((75-2x)/(2x))

x = 1.79

No of mol of HCl must add = salt = 2*1.79 = 3.58 mmol

volume of HCl must add = n/M = 3.58/12 = 0.3 ml

b) No of mol of NAoh added = 1*5 = 5 mmol

   No of mol of NaOH reacted with acidic form of buffer = 3.58 mmol

excess NaOH concentration = (5-3.58)/751 = 0.00189 M

   pH = 14 - (-log(OH^-))

       = 14 - (-log(0.00189))

       = 11.27

c) concentration of NaOH in the mixture = n/v

            = 1*5/751

            = 0.00666 M

      pH = 14 - (-log(OH^-))

       = 14 - (-log(0.00666))

    = 11.82