Question

a) For 50mL of a 0.400M borate buffer at pH 8.5 (boric acid Ka = 5.8x10^-10), Describe how you would make this using sodium borate (NaH2BO3) and 2.50M HCl.

b) what is the solution pH after adding 0.100 mL of concentrated HCl to a 20.0 mL portion of this buffer?

c) What is the solution pH after adding 10.0 mL of 6.00 M NaOH to a 20.0 mL portion of this original buffer?

Answer 1

a) No of mol of buffer = salt+acid = 50*0.4 = 20 mmol

   pH of acidic buffer = pka + log(salt/acid)

   pka of borate buffer = -log(5.8*10^-10) = 9.236

      8.5 = 9.236+logx

   x = 0.184

salt/acid = 0.184

   acid*0.184+acid = 20

    acid = x

     x*0.184+x = 20

No of mol of acid(HCl) = 16.89 mmol

volume of HCl required = n/M = 16.89/2.5 = 6.76 ml

No of mol of salt((NaH2BO3) = 20-16.89 = 3.11 mmol

Amount of sodium borate required = 3.11*10^-3*127.78 = 0.397 g

take 0.397 g of sodium borate and add 6.76 ml of HCl and add water upto 50 ml

b) No of mol of buffer = 20*0.4 = 8 mmol

pH of acidic buffer = pka + log(salt/acid)

   pka of borate buffer = -log(5.8*10^-10) = 9.236

      8.5 = 9.236+log(x/(8-x))

No of mol of sodium borate = x = 1.24 mmol

No of mol of acid = 8-x = 8-1.24 = 6.76 mmol

after addition of 0.1 ml HCl ,

No of mol of HCl added = V*M = 0.1*2.5 = 0.25 mmol

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

                = 9.236+log((1.24-0.25)/(6.76+0.25))

                = 8.386

c) i think its 1.00 ml 6 M NAoH

after addition of 1.00 ml 6 M NaOH ,

No of mol of acid present in buffer = 8-x = 8-1.24 = 6.76 mmol

No of mol of NaOH added = V*M = 1*6 = 6 mmol

pH of acidic buffer = pka + log(salt+NaoH/acid-NaOH)

                = 9.236+log((1.24+6)/(6.76-6))

                = 10.215