Question

Using the Henderson-Hasselbalch equation, calculate the amount of HEPES (sodium salt) and HEPES (free acid) required to prepare 50 ml of a 100 mM buffer that is pH = 7.20. The pKa of HEPES is 7.55 at 20° C. The formula weight of the sodium salt is 260.31. The formula weight of the free acid is 238.31. . Weigh out the appropriate amounts of the HEPES (sodium salt) and HEPES (free acid), transfer to a 100 mL beaker, dissolve in deionized water to an approximate volume of 40 mL

Answer 1

Explanation:

According to the Henderson-Hasselbalch equation,

              pH = $pK_{a} + log \frac{[Salt]}{[Acid]}$

where, [Salt] = molar concentration of sodium salt

           [Acid] = molar concentration of free acid

As per the given problem, pH = 7.20 and $pK_{a}$ = 7.55.

Hence, putting the given values into the above formula as follows.

                pH = $pK_{a} + log \frac{[Salt]}{[Acid]}$

                7.20 = 7.55 + $log \frac{Salt}{Acid}$

             $log \frac{Salt}{Acid}$ = -0.35

or,                  $\frac{Salt}{Acid}$ = 0.4467

Now, the concentration of buffer is 100 mM.

Hence,             [Salt] + [Acid] = 100 mM

              $[Acid](1 + \frac{[Salt]}{[Acid]}$ = 100 mM              

            [Acid](1 + 0.4467) = 100 mM

                    [Acid] = $\frac{100}{1.4467}$

                              = 69.123 mM

Therefore, concentration of salt will be calculated as follows.

                    [Salt] = 0.4467 × [Acid]

                             = $0.4467 \times 69.123 mM$

                             = 30.877 mM

For calculating the amount of acid:

     Desired concentration = 69.123 mM

So, in 1000 ml amount of acid should be $69.123 \times 10^{-3} mol$.

In 50 ml, the amount of acid should be as follows.

           $\frac{69.123 \times 10^{-3} \times 50}{1000} mol$

            = 0.824 g

or,        = 824 mg          (as 1 g = 1000 mg)

Calculation for the amount of sodium salt is as follows.

            Desired concentration = 30.877 mM

In 1000 ml, amount of salt should be $30.877 \times 10^{-3} mole$.

In 50 ml, amount of salt should be as follows.

        $\frac{30.877 \times 10^{-3} mole \times 50 \times 260.31 g}{1000}$

             = 0.402 g

or,          = 402 mg

Thus, we can conclude that we need to weigh out 824 mg of HEPES (free acid) and 402 mg of HEPES (sodium salt) that is required to be dissolved in 40 ml of water. After than, we have to adjust the volume of 50 ml by adding water drop-wise.