Respuesta :
Explanation:
According to the Henderson-Hasselbalch equation,
pH = [tex]pK_{a} + log \frac{[Salt]}{[Acid]}[/tex]
where, [Salt] = molar concentration of sodium salt
[Acid] = molar concentration of free acid
As per the given problem, pH = 7.20 and [tex]pK_{a}[/tex] = 7.55.
Hence, putting the given values into the above formula as follows.
pH = [tex]pK_{a} + log \frac{[Salt]}{[Acid]}[/tex]
7.20 = 7.55 + [tex]log \frac{Salt}{Acid}[/tex]
[tex]log \frac{Salt}{Acid}[/tex] = -0.35
or, [tex]\frac{Salt}{Acid}[/tex] = 0.4467
Now, the concentration of buffer is 100 mM.
Hence, [Salt] + [Acid] = 100 mM
[tex][Acid](1 + \frac{[Salt]}{[Acid]}[/tex] = 100 mM
[Acid](1 + 0.4467) = 100 mM
[Acid] = [tex]\frac{100}{1.4467}[/tex]
= 69.123 mM
Therefore, concentration of salt will be calculated as follows.
[Salt] = 0.4467 × [Acid]
= [tex]0.4467 \times 69.123 mM[/tex]
= 30.877 mM
For calculating the amount of acid:
Desired concentration = 69.123 mM
So, in 1000 ml amount of acid should be [tex]69.123 \times 10^{-3} mol[/tex].
In 50 ml, the amount of acid should be as follows.
[tex]\frac{69.123 \times 10^{-3} \times 50}{1000} mol[/tex]
= 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 [tex]30.877 \times 10^{-3} mole[/tex].
In 50 ml, amount of salt should be as follows.
[tex]\frac{30.877 \times 10^{-3} mole \times 50 \times 260.31 g}{1000}[/tex]
= 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.
