The pH of a weak monoprotic acid, HA, is 4.55. It took 39.22 ml of 0.2334 M NaOH to titrate 25.00 ml of the acid.
a. Write an equation for the above reaction.
b. Calculate the molarity of the weak acid
c. Write the equilibrium equation. Construct a table showing initial concentration, change in concentration and the equilibrium concentration. Calculate the ionization constant Ka, for the above acid
d. 23.55 ml of the NaOH were added to partially neutralize a new 25.00 ml sample of the acid. The pH of the mixture was measured as 5.33. Calculate Ka for the acid, HA, using the partial neautralization method
a) HA + NaOH ==> H2O + NaA
b) moles NaOH = M x L = ??
Using the coefficients in the balanced equation, convert moles NaOH to moles HA.
Then M HA = moles HA/L HA. You have moles and L, solve for HA.
c) Ka = (H^+)(A^-)/(HA)
millimoles HA = mL x M = 25.00*0.3662 = 9.154
mmoles NaOH = 9.154
I don't know what you want for the chart.
d)I don't know what the "partial neutralization" procedure is. I would use the Henderson-Hasselbalch equation.
..............HA + NaOH ==> NaA + H2O
I = initial. 25.00 x 0.3662M = mmoles.
add NaOH = 23.55 x 0.2334 = mmoles.
mmoles NaA = 23.55 x 0.2334
mmoles HA = initial mmoles - mmoles NaOH . Then pH = 5.33 = pKa + log(base/acid) and solve for pKa, then Ka.
a. The equation for the reaction between the weak monoprotic acid (HA) and NaOH is:
HA + NaOH -> H2O + NaA
b. To calculate the molarity of the weak acid, we can use the formula M1V1 = M2V2.
Given:
M1 (NaOH) = 0.2334 M
V1 (NaOH) = 39.22 mL
M2 (HA) = ?
V2 (HA) = 25.00 mL
Using the formula:
(0.2334 M)(39.22 mL) = M2 (25.00 mL)
M2 = (0.2334 M)(39.22 mL) / 25.00 mL
M2 ≈ 0.3669 M
Therefore, the molarity of the weak acid (HA) is approximately 0.3669 M.
c. The equilibrium equation for the ionization of the weak monoprotic acid (HA) can be represented as follows:
HA ⇌ H+ + A-
Constructing a table:
| Species | Initial Concentration | Change in Concentration | Equilibrium Concentration |
|---------------------|---------------------------|-----------------------------|--------------------------------|
| HA | 0.3669 M | -x | 0.3669 - x M |
| H+ | 0 | +x | x M |
| A- | 0 | +x | x M |
Since the initial concentration of HA is equal to its equilibrium concentration (0.3669 M), we can ignore the change in concentration for this species.
Using the equation for Ka:
Ka = (H+)(A-) / (HA)
Ka = (x)(x) / (0.3669 - x)
d. Given that the pH of the mixture (HA + NaOH) is 5.33, we can convert this value to [H+] using the equation:
[H+] = 10^(-pH)
[H+] = 10^(-5.33)
Using the equation for Ka:
Ka = (H+)(A-) / (HA)
Ka = (10^(-5.33))(x) / (0.3669 - x)
The volume of NaOH used to partially neutralize the new 25.00 mL sample is 23.55 mL, which is similar to the V2 in part b of the question.
We can assume x represents the concentration of H+ and A-. Therefore, x = 10^(-5.33).
Substituting the values into the equation:
Ka = (10^(-5.33))(10^(-5.33)) / (0.3669 - 10^(-5.33))
Calculating the value gives us the Ka for the acid, HA, using the partial neutralization method.
a. The equation for the reaction between the weak monoprotic acid, HA, and the sodium hydroxide, NaOH, is:
HA + NaOH → H2O + NaA
b. To calculate the molarity of the weak acid, we can use the formula:
Molarity (M) = moles of solute / volume of solution (in liters)
In the titration, the volume of NaOH used was 39.22 mL, which is 0.03922 L. The molarity of NaOH is 0.2334 M. From the balanced equation, we can see that the mole ratio between HA and NaOH is 1:1. Therefore, the moles of HA can be calculated as:
moles of HA = moles of NaOH = Molarity x Volume = 0.2334 M x 0.03922 L = 0.009152 moles
Since the volume of the acid used was 25.00 mL, which is 0.02500 L, the molarity of the weak acid can be calculated as:
Molarity of HA = moles of HA / volume of HA = 0.009152 moles / 0.02500 L = 0.3661 M
So, the molarity of the weak acid, HA, is 0.3661 M.
c. The equilibrium equation for the ionization of the weak acid, HA, can be written as:
HA (aq) ⇌ H+ (aq) + A- (aq)
Using this equilibrium equation, we can construct a table to show the initial concentration, change in concentration, and equilibrium concentration:
| | HA (aq) | H+ (aq) | A- (aq) |
|--------|-----------|------------|-----------|
| Initial| 0.3661 M | 0 M | 0 M |
| Change | -x | +x | +x |
| Equil | 0.3661-x | x | x |
The ionization constant, Ka, can be calculated using the equation:
Ka = [H+][A-] / [HA]
At equilibrium, the concentrations are:
[H+] = x
[A-] = x
[HA] = 0.3661 - x
Substituting these values into the equation for Ka:
Ka = (x)(x) / (0.3661 - x)
d. To calculate Ka using the partial neutralization method, we can use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA])
In this case, the pH is 5.33. Since pH = -log[H+], we can convert the pH to [H+] concentration:
[H+] = 10^(-pH) = 10^(-5.33) = 5.01 x 10^(-6) M
From the equilibrium table, we know that [HA] = 0.3661 M and [A-] = x.
Substituting the values into the Henderson-Hasselbalch equation:
5.33 = -log(Ka) + log(x / 0.3661)
Rearranging the equation:
log(Ka) = -5.33 + log(x / 0.3661)
Ka = 10^(-5.33 + log(x / 0.3661))
To solve for Ka, we need to find the value of [A-] (or x) from the equilibrium equation. Given that 23.55 mL of NaOH were added, which is 0.02355 L, and the molarity of NaOH is 0.2334 M, we can calculate the moles of NaOH used:
moles of NaOH = Molarity x Volume = 0.2334 M x 0.02355 L = 0.005492 moles
Since the mole ratio between NaOH and HA is 1:1, this is also the moles of HA that were neutralized. From the initial moles of HA, we can subtract the moles neutralized to find the remaining moles of HA:
moles of HA remaining = initial moles of HA - moles neutralized
= 0.009152 moles - 0.005492 moles
= 0.00366 moles
The equilibrium concentration [A-] can be calculated using the equation:
[A-] = moles / volume = 0.00366 moles / 0.02500 L = 0.1464 M
Now we can substitute this value back into the Henderson-Hasselbalch equation to solve for Ka:
Ka = 10^(-5.33 + log(0.1464 / 0.3661))
Calculating this expression will give the value of Ka for the acid, HA, using the partial neutralization method.
a. The equation for the reaction between the weak acid, HA, and the strong base, NaOH, can be written as:
HA + NaOH → NaA + H2O
b. To calculate the molarity of the weak acid, we need to use the equation for neutralization reaction:
Moles of acid = Moles of base
First, we need to calculate the moles of NaOH used:
Moles of NaOH = volume (in L) × concentration (in M)
= 39.22 mL × 0.2334 M / 1000 mL/L
Using this value, we can now find the moles of acid:
Moles of acid = Moles of NaOH
Since we know the volume of the acid is 25.00 mL, we convert it to liters:
Volume of acid = 25.00 mL / 1000 mL/L
Finally, we can calculate the molarity of the weak acid:
Molarity of acid = Moles of acid / Volume of acid
c. The equilibrium equation for the dissociation of the acid, HA, is:
HA ⇌ H+ + A-
To analyze the equilibrium, we can construct a table:
HA ⇌ H+ + A-
Initial [HA]0 0 0
Change -x +x +x
Equilibrium [HA]0 - x x x
The ionization constant, Ka, for the acid can be calculated using the equation:
Ka = [H+] [A-] / [HA]
Substituting the equilibrium concentrations into the equation, we get:
Ka = (x) (x) / ( [HA]0 - x )
d. The partial neutralization method involves measuring the pH of a mixture of the weak acid with a known volume of the base.
First, calculate the moles of NaOH used in the partial neutralization:
Moles of NaOH = volume (in L) × concentration (in M)
= 23.55 mL × 0.2334 M / 1000 mL/L
Since we know the volume of the acid is 25.00 mL, we convert it to liters and calculate the moles of acid:
Moles of acid = Moles of NaOH
Using the pH of the mixture, we can calculate the concentration of H+ ions:
[H+] = 10^(-pH)
Substituting the values into the equation for the ionization constant, Ka:
Ka = [H+] [A-] / [HA]
We can assume that the concentration of A- is equal to the concentration of NaOH used since it is a monoprotic acid. Now we solve for Ka using the calculated concentrations.