You wish to prepare 800 mL of a buffer solution at pH 5.10 that utilizes the ace

Question

You wish to prepare 800 mL of a buffer solution at pH 5.10 that utilizes the acetic acid/acetate equilibrium and has a total acetate concentration (acetic acid plus acetate concentrations) of 1.00 M. In your lab, you discover that you don't have any aceitc acid, but you do have a large bottle of sodium acetate dihydrate, NaC2H3O2*2H2O. You also have a bottle of 1.50 M NaOH. You know that you can make the buffer by dissolving the correct mass of sodium acetate dihydrate in water and then adding the appropriate volume of NaOH before diluting to final volume. How many mL of the 1.50M NaOH solution are required to be added to the solution during preparation to attain an expected pH of 5.10?

Explanation / Answer

pH of acidic buffer = 5.1

pH = pka + log(salt/acid)

5.1 = 4.74 + log(salt/acid)

(salt/acid) = 10^0.36 = 2.29 ------- 1

salt+acid = (800/1000)*1 = 0.8 ------- 2

2.29acid + acid = 0.8

acid = 0.243

base = 0.8-0.243 = 0.557

take 0.243+0.557 = 0.8 mole of (NaC2H3O2*2H2O)

amount of NaC2H3O2*2H2O) must be taken = 0.8*118

                       = 94.4 grams

No of moles of NaOH must be taken = V*1.5 = 0.557

V = volume of NaOH = 0.557/1.5 = 0.371 L

V =    volume of NaOH = 371 ml

so take 94.4 grams of NaC2H3O2*2H2O , 371 ml volume of NaOH and make the solution up to

800 ml solution

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.