Calculate the volume of 0.05 M NaOH which must be added to 100 cm3 of 0.05 M methanoic acid to obtain a buffer of pH 4.23.�(Ka for methanoic acid = 1.77 x 10-4)
methanoic acid = HM
millimols HM = 100 x 0.05 = 5.
...HM + OH^- ==> M^- + H2O
I...5....0.......0.....
add......x...........
C..-x...-x.......x
E..5-x...x.......x
Substitute the E line into the Henderson-Hasselbalch equation and solve for x = millimols NaOH.
Then M NaOH = mmols NaOH/mL NaOH. Substitute and solve for mL of 0.05 M NaOH. I've estimated the answer at about 25 mL of the .05 M NaOH. Post your work if you get stuck.
To calculate the volume of 0.05 M NaOH needed to obtain a buffer of pH 4.23, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer to the pKa of the acid and the ratio of conjugate base to acid.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([conjugate base] / [acid])
In this case, the acid is methanoic acid (HCOOH), and its conjugate base is the formate ion (HCOO-).
First, let's find the pKa of methanoic acid using the given Ka value:
Ka = [HCOO-][H+]/[HCOOH]
Assuming that the initial concentration of methanoic acid is equal to its final concentration in the buffer solution, we can substitute the values:
1.77 x 10^(-4) = [HCOO-][H+]/[HCOOH]
Since NaOH is a strong base, it will dissociate completely in water, giving us Na+ and OH- ions. Therefore, the concentration of hydroxide ions available for reaction is the same as the concentration of NaOH.
Let's assume x represents the amount (in moles) of NaOH required to react with the same amount of methanoic acid. Then the final concentrations of HCOO- and HCOOH will be (0.05 - x) M and (0.05 + x) M, respectively.
Now we can substitute these values into the Henderson-Hasselbalch equation:
4.23 = -log(1.77 x 10^-4) + log((0.05 - x) / (0.05 + x))
To solve for x, we can rearrange the equation:
3.77 = log((0.05 - x) / (0.05 + x))
10^(3.77) = (0.05 - x) / (0.05 + x)
10^(3.77) * (0.05 + x) = 0.05 - x
0.05 + 10^(3.77) * (0.05 + x) = 0.05 - x
0.05 + 0.05 * 10^(3.77) + x * 10^(3.77) = 0.05 - x
0.05 * 10^(3.77) + x * 10^(3.77) = -0.05 - x
(0.05 * 10^(3.77)) / (10^(3.77) + 1) = -0.05 / (10^(3.77) + 1)
Solving this equation gives:
x ≈ 0.0039
So, approximately 0.0039 moles (or 0.0039 * 1000 = 3.9 cm^3) of NaOH is needed to obtain the desired buffer solution.
To calculate the volume of NaOH needed to form a buffer of pH 4.23, first, we need to determine the concentration of the methanoic acid in the buffer solution.
We know that the concentration of the methanoic acid (HCOOH) is 0.05 M. In a buffer solution, the pH is determined by the ratio of the concentration of the acid to its conjugate base (in this case, HCOO-).
Using the Henderson-Hasselbalch equation, pH = pKa + log([base]/[acid]), we can rearrange the equation to solve for the ratio [base]/[acid].
pH = 4.23 (given)
pKa = -log(Ka) = -log(1.77 x 10^(-4))
[base]/[acid] = 10^(pH - pKa)
[base]/[acid] = 10^(4.23 - (-log(1.77 x 10^(-4))))
Now, let's calculate the ratio [base]/[acid]:
[base]/[acid] = 10^(4.23 - (-log(1.77 × 10^(-4))))
[base]/[acid] = 10^(4.23 + 4.75)
[base]/[acid] = 10^(9.98)
[base]/[acid] = 9.78 × 10^9
Since the concentration of the methanoic acid is 0.05 M, we can multiply the concentration of the acid by the calculated ratio to find the concentration of the base (NaOH).
[base] = 9.78 × 10^9 × [acid]
[base] = 9.78 × 10^9 × 0.05 M
Now, to find the volume of 0.05 M NaOH needed, we can use the formula:
volume = [base] / concentration of NaOH
volume = (9.78 × 10^9 × 0.05 M) / 0.05 M
volume = 9.78 × 10^9 cm^3
Therefore, the volume of 0.05 M NaOH that must be added to 100 cm^3 of 0.05 M methanoic acid to obtain a buffer with pH 4.23 is 9.78 × 10^9 cm^3.