Question

In: Chemistry

You need to prepare 100.0 mL of a pH=4.00 buffer solution using 0.100 M benzoic acid...

You need to prepare 100.0 mL of a pH=4.00 buffer solution using 0.100 M benzoic acid (pKa = 4.20) and 0.220 M sodium benzoate. How much of each solution should be mixed to prepare this buffer?

a) mL of benzoic acid

b) mL of sodium benzoate

Solutions

Expert Solution

Let volume of sodium benzoate =x

      volume of benzoic acid      = 100-x

      no of moles of sodium benzoate = molarity * volume

                                                     = 0.22*x

     no of moles of bezoic acid            = molarity * volume

                                                       = 0.1*(100-x)

       PH = PKa +log[sodium benzoate]/[benzoic acid]

       4      = 4.2 + log0.22*x/0.1*(100-x)

       4-4.2   = log0.22*x/0.1*(100-x)

       -0.2     = log0.22*x/0.1*(100-x)

log0.22*x/0.1*(100-x) = -0.2

0.22*x/0.1*(100-x)       = 10-0.2 = 0.63

0.22*x/0.1*(100-x)   = 0.63

0.22*x               = 0.63*0.1(100-x)

                             = 0.063(100-x)

0.22x                    =6.3-0.063x

0.22x-6.3+0.063x =0

0.263x-6.3 =0

   0.263x = 6.3

         x   = 6.3/0.263   = 23.95

volume of benzoic acid   = 23.95ml

volume of benzoate       = 100-x = 100-23.95 = 76.05ml

   

      

queen_honey_blossom answered 2 years ago
Next > < Previous

Related Solutions

You need to prepare 100.0 mL of a pH=4.00 buffer solution using 0.100 M benzoic acid...
You need to prepare 100.0 mL of a pH=4.00 buffer solution using 0.100 M benzoic acid (pKa = 4.20) and 0.120 M sodium benzoate. How much of this solution should be mixed to prepare this buffer? Sum of volumes must equal 100mL. Please explain clearly and show work! Thank you :)
You need to prepare 100.0 mL of a pH = 4.00 buffer solution using 0.100 M...
You need to prepare 100.0 mL of a pH = 4.00 buffer solution using 0.100 M benzoic acid (pKa = 4.20) and 0.200 M sodium benzoate. How much of each solution should be mixed to prepare this buffer?
How can you prepare 100.0 mL of acetic buffer solution (0.100 M) with pH = 5.10...
How can you prepare 100.0 mL of acetic buffer solution (0.100 M) with pH = 5.10 by using a solution of acetic acid (25% w/w, density = 1.05 g/cm3 ) and pure sodium acetate. (pKa = 4.76; M. W. = 60.05 g/mol for HAc and 82.03 for NaAc).
Calculate the change in pH when 4.00 mL of 0.100 M HCl(aq) is added to 100.0...
Calculate the change in pH when 4.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq). Delta pH= ? Calculate the change in pH when 4.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution. Delta pH= ?
What is the pH for a 400 ml solution of 1.6 M Benzoic Acid? If you...
What is the pH for a 400 ml solution of 1.6 M Benzoic Acid? If you added 75 grams of conjugate base (NaC6H5CO2) to the solution, what would the new pH be? If you now added 75 ml of 1.35 M NaOH, what would be the new pH?
1)How many grams of KH2PO4 are needed to prepare 100.0 mL of a 0.100 M buffer...
1)How many grams of KH2PO4 are needed to prepare 100.0 mL of a 0.100 M buffer with a pH of 7.25 (assume a pKa of 6.86)? 2)How many grams of K2HPO4 are needed to prepare 100.0 mL of a 0.100 M buffer with a pH of 6.75 (assume a pKa of 6.86)? 3) Phosphoric acid has three pKa’s (pKa1, pKa2, pKa3), which pKa is most relevant to the buffer in an experiment?
A 500.0 mL buffer solution is 0.10 M in benzoic acid and 0.10 M in sodium...
A 500.0 mL buffer solution is 0.10 M in benzoic acid and 0.10 M in sodium benzoate and has an initial pH of 4.19. What is the pH of the buffer upon addition of 0.020 mol of NaOH?
You prepare a buffer solution from 10.0 mL of 0.100 M MOPS (3-morpholinopropane-1-sulfonic acid) and 10.0...
You prepare a buffer solution from 10.0 mL of 0.100 M MOPS (3-morpholinopropane-1-sulfonic acid) and 10.0 mL of 0.085 M NaOH. Next, you add 1.00 mL of 1.71 × 10-5 M lidocaine to this mixture Denoting lidocaine as L, calculate the fraction of lidocaine present in the form LH .
You prepare a buffer solution from 10.0 mL of 0.100 M MOPS (3-morpholinopropane-1-sulfonic acid) and 10.0...
You prepare a buffer solution from 10.0 mL of 0.100 M MOPS (3-morpholinopropane-1-sulfonic acid) and 10.0 mL of 0.079 M NaOH. Next, you add 1.00 mL of 1.65 × 10-5 M lidocaine to this mixture Denoting lidocaine as L, calculate the fraction of lidocaine present in the form LH . ka=6.3e-8 kb=8.7e-7
You prepare a buffer solution from 10.0 mL of 0.100 M MOPS (3-morpholinopropane-1-sulfonic acid) and 10.0...
You prepare a buffer solution from 10.0 mL of 0.100 M MOPS (3-morpholinopropane-1-sulfonic acid) and 10.0 mL of 0.079 M NaOH. Next, you add 1.00 mL of 5.67 × 10^-5 M lidocaine to this mixture. Denoting lidocaine as L, calculate the fraction of lidocaine present in the form LH+. MOPS Ka = 6.3 × 10^–8. Lidocaine Kb = 8.7 × 10^–7 (Hint: First calculate the pH of the solution from the amount of MOPS added. Start by finding the number...
ADVERTISEMENT
Subjects
ADVERTISEMENT
Latest Questions
ADVERTISEMENT