Question

Bromophenol Blue is often used as an acid-base indicator. In neutral and basic conditions it is blue while in acidic conditions it is yellow. The yellow form of bromophenol blue absorbs at 440nm with a molar absorptivity of 5.85x10^4 (assume that the blue form does not absorb at this wavelength). The pKa of bromophenol blue is 4. you would like to make 3ml solution of bromophenol blue that has an absorbance at440= 0.560. Select the buffer that you would use to make the solution: ANSWER CHOICES A: 50mM phosphate buffer pH7.6, B: 100mM glycine buffer pH 2.0 C: 50mM sodium acetate pH 4.2 D:50mM phosphate buffer pH 5.7. You have a 5.5x10^-4M bromophenol blue solution. What volume of this solution will you use to make your solution (3mL of Abs440=0.56 solution)?   

Answer 1

1. Select the buffer that you would use to make the solution:

    The pKa of bromophenol blue is 4. Best choice for pH of buffered solution is close to pKa value. Thus choice shall be :   C: 50mM sodium acetate pH 4.2

It is the only buffer in range of pKa for bromophenol blue.

2. What volume of this solution will you use to make your solution (3mL of Abs440=0.56 solution)?   

   We have a 5.5x10^-4M bromophenol blue solution.

• Bromophenol blue absorbs at 440nm with a molar absorptivity : 5.85*10^4 M^-1cm^-1

  we have, pKa is bromophenol blue = 4.0

  The yellow (protonated) form of bromophenol blue absorbs , thus yellow (protonated) form of bromophenol : HA

  and bromophenol blue : A.

  from, Beer's law , A = e*C*l

  we need A(440) = 0.56      ; l = 1 cm (usually).

  thus, 0.56 = 5.85*10^4 M^-1cm^-1 * C * 1cm  

  C = 9.57*10^-6 M

  Thus, we need [HA] = 9.57*10^-6 M

  from, Henderson-Hasselbalch equation for pH :

  pH = pKa + log {[A] / [HA]}

  pH = 4.2 (buffer used)

  log {[A] / [HA]} = 0.2

  or [A] / [HA] = 1.6

  since, [HA] = 9.57*10^-6 M

  thus, [A] = 1.6* [HA] = 1.52*10^-5 M

  thus, [A] +[HA] = 2.5*10^-5 M

  we need 3 ml solution,

  thus , Volume of 5.5 x 10^-4 M bromophenol blue solution needed :

  ( 3 mL *2.5*10-5 mol/L )/5.5 x 10^-4 mol/L   = 0.135 mL