15. (23 points total) Show calculations for partial credit. A buffer is made fro

Question

15. (23 points total) Show calculations for partial credit. A buffer is made from formic acid (HCHO2) and sodium formate (NaCHO.). The buffer has 0.2 moles of HCHO2 per liter of solution and 0.15 moles of NaCHO2 again per liter of solution. For HCHO2 K 1.8 x 104 7 points) What is the initial pH of this buffer system? (b) (8 points)What is the resulting pH after the addition of 1 mL of 0.100 M NaOH to one liter of this buffer system? (c) (8 points) What is the resulting pH when one liter of the original buffer solution has 1.5 ml of 0.100 M HCI added to it?

Explanation / Answer

a) pH of acidic buffer = pka + log(conjugate base/acid)

   [base] = [NaCHO2] = 0.15 M

   [acid] = [HCHO2] = 0.2 M

   pka of weak acid = -logka

                    = -log(1.8*10^-4)

                    = 3.74

pH = 3.74 + log(0.15/0.2)

     = 3.61506

b) no of mol of NaOH added = 1*0.1/1000 = 0.0001 mol

   pH of acidic buffer = pka + log(conjugate base + NaOH/acid - NaOH)

pH = 3.74 + log((0.15+0.0001)/(0.2-0.0001))

      = 3.6156

c) no of mol of HCl added = 1.5*0.1/1000 = 0.00015 mol

   pH of acidic buffer = pka + log(conjugate base + HCl/acid - HCl)

pH = 3.74 + log((0.15-0.00015)/(0.2+0.00015))

     = 3.6143

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