Question

How many liters of 0.250 M of NaOH are needed to create 750.0 mL of a 100.00 mM histidine buffer at pH=2.50?

If you added 1.00 mL of a 6.00 M HCl to the buffer (750.0 mL) in the part above, what would be the resulting pH? If you added 1.00 mL of a 6.00 M HCl to 750 mL of pure water what would be the pH of this solution be? CIRCLE BOTH ANSWERS

Answer 1

pH of histidine buffer = pka1 + log(salt/acid)

pka1 of histidine = 1.82

no of mol of histidine taken = 750*100*10^-3 = 75 mmol

no of mol of NaOH must be added = x

2.5 = 1.82 + log(x/(75-x))

x = 62 mmol

volume NaOH must be added = 62/0.25 = 248 ml

b) no of mol of HCl added to buffer = 1*6 = 6 mmol

pH of histidine buffer = pka1 + log(salt - HCl/acid+HCl)

     = 1.82 + log((62-6)/(75-62+6))

     = 2.29

c) concentration of HCl in water = M*v/Vtotal

                = 6*1/751 = 0.008 M

   pH = -log(H3O+)

       = -LOG0.008

       = 2.1