Find the ratio of the volumes of NaC3H5O3 soloution to HC3H5O3 (lactic acid) required to prepare a buffer with a pH of 3.90. the solutions are equal in molarity. Ka = 1.4 x 10-4 for HC3H5O3.
ratio = ________ mL of NaC3H5O3 per mL of HC3H5O3
To find the ratio of volumes of NaC3H5O3 solution to HC3H5O3 required to prepare a buffer with a pH of 3.90, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
Where pH is the desired pH of the buffer, pKa is the acid dissociation constant, [A-] is the concentration of the conjugate base (NaC3H5O3), and [HA] is the concentration of the acid (HC3H5O3).
Since the solutions are equal in molarity, we can express the concentrations as:
[A-] = [HA]
Let's assume the initial concentration is 'C'.
So, [A-] = C and [HA] = C.
Also, we can convert the pKa to Ka:
Ka = 10^(-pKa)
Therefore, Ka = 10^(-4.90) = 1.4 x 10^(-5)
Now, let's substitute the values into the Henderson-Hasselbalch equation:
3.90 = -log(1.4 x 10^(-5)) + log(C/C)
Simplifying the equation:
3.90 = -(-4.85) + log(1)
3.90 + 4.85 = log(1) + log(C/C)
8.75 = log(C/C)
Now, let's convert the logarithmic equation into an exponential equation:
10^8.75 = C/C
Simplifying:
C^2 = 10^8.75
Taking the square root:
C = sqrt(10^8.75)
C ≈ 5.623 x 10^4
Therefore, the concentration of both NaC3H5O3 and HC3H5O3 is approximately 5.623 x 10^4 M.
To find the ratio of volumes, we need to calculate the moles of NaC3H5O3 and HC3H5O3 required.
Moles = Concentration x Volume
Let's assume the volume of HC3H5O3 is 'V'.
Moles of HC3H5O3 = (5.623 x 10^4) x V
To maintain the equal molarities, the volume of NaC3H5O3 will also be 'V'.
Moles of NaC3H5O3 = (5.623 x 10^4) x V
The ratio of volumes of NaC3H5O3 solution to HC3H5O3 is:
Volume ratio = Volume of NaC3H5O3 / Volume of HC3H5O3
= (5.623 x 10^4) x V / V
= 5.623 x 10^4
Therefore, the ratio is approximately 5.623 x 10^4 mL of NaC3H5O3 per mL of HC3H5O3.
To find the ratio of the volumes of NaC3H5O3 solution to HC3H5O3 required to prepare the buffer, we need to use the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation relates the pH of a buffer solution to the pKa of the acidic component and the ratio of its concentrations.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log ([A-] / [HA])
In this case, NaC3H5O3 is the conjugate base (A-) and HC3H5O3 is the acid (HA) in the buffer solution. We are given that the pH of the buffer is 3.90 and the pKa of HC3H5O3 is 1.4 x 10^-4.
Rearranging the equation, we have:
log ([A-] / [HA]) = pH - pKa
Taking the antilog of both sides, we have:
[A-] / [HA] = 10^(pH - pKa)
Since the solutions are equal in molarity, we can substitute [A-] / [HA] with the ratio of their volumes. Let's call the volume of NaC3H5O3 solution V_Na and the volume of HC3H5O3 solution V_H.
(V_Na / V_H) = 10^(pH - pKa)
Now, let's plug in the values:
pH = 3.90
pKa = 1.4 x 10^-4
(V_Na / V_H) = 10^(3.90 - 1.4 x 10^-4)
Calculating this on a calculator, we find that the ratio of the volumes of NaC3H5O3 solution to HC3H5O3 required to prepare the buffer is approximately:
ratio = 244657 : 1 (rounded to the nearest whole number)
Therefore, the ratio is 244657 mL of NaC3H5O3 per mL of HC3H5O3.