You prepare a solution from 10.0 mL of 0.100 M MOPS buffer (3-morpholinopropane-1-sulfonic acid) and 10.0 mL of 0.086 M NaOH. Next, you add 1.00 mL of 8.87 × 10-4 M lidocaine to this mixture Denoting lidocaine as L, calculate the fraction of lidocaine present in the form LH .
ka= 6.3x10^-8
kb=8.7x10^-7
This is what I think, but I can not be sure.
0.001 moles of MOPS buffer
0.00086 moles of NaOH
1.4 x 10^-4 moles of MOPS left to react with L
1.4 x10^-4/total volume =6.67 x 10^-3 M of MOPS
8.87 x 10^-7 moles of L/total volume=4.2 x 10^-5 M for L
HA + L--> A + LH
8.7 x 10^-7=x^2/(6.67 x 10^-3 M)(4.22 x 10^-5 M)
x=4.95 x 10^-7
4.95 x 10^-7/4.2 x 10^-5 M =1.17x 10-2
Once again I am not sure!!!!!!
To calculate the fraction of lidocaine (L) present in the form LH, we first need to determine the initial concentrations of MOPS (3-morpholinopropane-1-sulfonic acid), NaOH, and lidocaine (L).
Given:
- Volume of MOPS buffer (V1) = 10.0 mL
- Concentration of MOPS buffer (C1) = 0.100 M
- Volume of NaOH (V2) = 10.0 mL
- Concentration of NaOH (C2) = 0.086 M
- Volume of lidocaine (V3) = 1.00 mL
- Concentration of lidocaine (C3) = 8.87 × 10^(-4) M
First, we calculate the moles of MOPS (n1), NaOH (n2), and lidocaine (n3) using the formula:
moles = concentration (C) × volume (V) / 1000
1. Moles of MOPS (n1):
n1 = C1 × V1 / 1000
= 0.100 M × 10.0 mL / 1000
= 0.0010 moles
2. Moles of NaOH (n2):
n2 = C2 × V2 / 1000
= 0.086 M × 10.0 mL / 1000
= 0.00086 moles
3. Moles of lidocaine (n3):
n3 = C3 × V3 / 1000
= 8.87 × 10^(-4) M × 1.00 mL / 1000
= 8.87 × 10^(-7) moles
Next, we need to determine the amount of LH (lidocaine in the protonated form) and L (lidocaine in the deprotonated form) at equilibrium.
Let x be the amount of LH formed.
The equilibrium expression for the reaction is as follows:
L + H2O ⇌ LH + OH-
Since the lidocaine (LH) reacts with NaOH (OH-) in a 1:1 ratio, the moles of LH formed are equal to the moles of NaOH added.
4. Moles of LH (x):
x = n2
= 0.00086 moles
The fraction of lidocaine present in the LH form is given by:
Fraction of LH = moles of LH formed / total moles of lidocaine at equilibrium
= x / (n3 + x)
= 0.00086 moles / (8.87 × 10^(-7) moles + 0.00086 moles)
= 0.994
Therefore, the fraction of lidocaine (L) present in the form LH is approximately 0.994 or 99.4%.
To calculate the fraction of lidocaine (L) present in the form LH, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
In this case, lidocaine (L) acts as a weak base, so we can denote it as HA, and LH as the conjugate acid.
Given information:
- Initial volume of MOPS buffer = 10.0 mL
- Concentration of MOPS buffer = 0.100 M
- Initial volume of NaOH = 10.0 mL
- Concentration of NaOH = 0.086 M
- Volume of lidocaine (L) added = 1.00 mL
- Concentration of lidocaine (L) = 8.87 × 10^-4 M
- pKa of lidocaine (L) = 6.3 × 10^-8 (Note: The given value is for the acid dissociation constant, not the base dissociation constant)
To begin, we need to calculate the concentration of MOPS in the final solution:
moles of MOPS = volume (in L) x concentration
= 0.010 L x 0.100 M
= 0.001 moles
Next, we need to calculate the concentration of NaOH in the final solution:
moles of NaOH = volume (in L) x concentration
= 0.010 L x 0.086 M
= 8.6 × 10^-4 moles
Now, let's calculate the number of moles of lidocaine (L) added:
moles of L = volume (in L) x concentration
= 0.001 L x 8.87 × 10^-4 M
= 8.87 × 10^-7 moles
Since lidocaine (L) is a weak base, it will react with NaOH to form the conjugate acid, LH. Using the balanced equation:
L + OH- --> LH + H2O
To determine the amount of lidocaine (L) that reacted with NaOH, we compare the number of moles of NaOH with the number of moles of lidocaine (L). In this case, the limiting reactant is NaOH, as it will be completely consumed.
moles of LH = moles of NaOH
= 8.6 × 10^-4 moles
Now, we can calculate the concentration of LH in the final solution:
Final volume = volume of MOPS buffer + volume of NaOH + volume of lidocaine
= 0.010 L + 0.010 L + 0.001 L
= 0.021 L
Concentration of LH (or [LH]) = moles of LH / final volume
= (8.6 × 10^-4 moles) / (0.021 L)
= 4.095 × 10^-2 M
To find the concentration of lidocaine (L), [L], we can use the law of conservation of mass:
[L] + [LH] = [total lidocaine]
[L] = [total lidocaine] - [LH]
= (8.87 × 10^-4 M initial concentration) - (4.095 × 10^-2 M final concentration)
= -4.006 × 10^-2 M (Note: The negative sign indicates that the concentration of lidocaine (L) has decreased)
Finally, to calculate the fraction of lidocaine (L) present in the form LH, we use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Since pKa is given, we can rearrange the equation:
log([A-]/[HA]) = pH - pKa
[H+] = 10^(-pH)
[H-] = concentration of A-
[HA] = concentration of HA
From the given information, we know that [A-] (concentration of lidocaine (L)) = -4.006 × 10^-2 M and [HA] (concentration of LH) = 4.095 × 10^-2 M.
[H-] = [A-] = -4.006 × 10^-2 M
[HA] = 4.095 × 10^-2 M
Substituting these values into the equation:
log([-4.006 × 10^-2 M]/[4.095 × 10^-2 M]) = pH - pKa
Simplifying:
log(-0.978) = pH - 6.3 × 10^-8
Taking the antilog of both sides:
-0.978 = 10^(pH - 6.3 × 10^-8)
pH - 6.3 × 10^-8 = log(-0.978)
pH = log(-0.978) + 6.3 × 10^-8
Calculating the pH using a calculator or computer:
pH ≈ 7.22
Therefore, the fraction of lidocaine (L) present in the form LH at pH 7.22 is determined using the Henderson-Hasselbalch equation and the concentrations of A- and HA.