A student was required to prepare 250.0 mL of a hypochlorous acid/sodium hypochlorite buffer in which the concentration of the weak acid component was 0.073 M and the concentration of the conjugate base was 0.046 M. The student was supplied with 0.365 M hypochlorous acid and 1.0M NaOH to perform this task. What volume (in L) of the acid would the student need to prepare this buffer solution?
M = mols/L. You want 0.073M HClO and you want 0.250 L so that is 0.073 = mols/0.250 so mols = approx 0.01825 (I am using more significant figures than allowd---a few guard digits). Basically you want 18.25 mmols HClO in the solution.
For the base, you want mmols 250 x 0.046 = about 11.5 mmols NaClO.
We want to know how much HClO to start with.
.........HClO + NaOH ==> NaClO + H2O
I.........X......0.........0
add..............y.........y
C........-y......-y........y
E.......X-y......0.........y
We know y = 11.5 mmols NaOH to produce 11.5 mmols NaClO. If X-y = 18.2 then X = 18.25+y or 18.25+11.5 = 29.75 mmols HClO we must start with.
Since M = mmols/mL and we want 29.75 mmols, we must mL of the 0.365 stuff = 29.75/0.365 = about 81.5 mL of the 0.365 M stuff. You can check that out this way.
You will take 81.5 mL of 0.365M HClO and add 11.5 mL of the 1M NaOH. That will produce 11.5 mmols NaClO and leave (81.5 x 0.365 = 29.75)-11.5 = 18.25 mmols HClO
(HClO) = 18.25/250 = 0.073M which is what you wanted.
(NaClO) = 11.5/250 = 0.046 which is what you wanted.
To find the volume of the hypochlorous acid (HOCl) that the student needs to prepare the buffer solution, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
First, let's calculate the pKa of the hypochlorous acid. The pKa is the negative logarithm of the acid dissociation constant (Ka). In this case, the pKa of HOCl is 10.22.
Next, we can calculate the ratio of the concentrations of the conjugate base (A-) to the weak acid (HA) using the given concentrations:
[A-]/[HA] = 0.046 M / 0.073 M = 0.63
Using the Henderson-Hasselbalch equation, we can rearrange it to solve for [A-]/[HA]:
0.63 = 10.22 + log([A-]/[HA])
Subtracting 10.22 from both sides:
0.63 - 10.22 = log([A-]/[HA])
-9.59 = log([A-]/[HA])
Now, we can calculate the antilog of -9.59 to find the ratio [A-]/[HA]:
[A-]/[HA] = 10^(-9.59) = 1.14 x 10^(-10)
Since the ratio [A-]/[HA] represents the concentration of the conjugate base (0.046 M) over the concentration of the weak acid (0.073 M), we can set up the following equation:
0.046 M / (0.073 M - x) = 1.14 x 10^(-10)
Where x is the volume of the hypochlorous acid (in liters) that needs to be added to the solution.
Rearranging the equation:
0.046 M = (0.073 M - x) * 1.14 x 10^(-10)
0.046 M = 0.073 M * 1.14 x 10^(-10) - x * 1.14 x 10^(-10)
x * 1.14 x 10^(-10) = 0.073 M * 1.14 x 10^(-10) - 0.046 M
x * 1.14 x 10^(-10) = 5.06 x 10^(-12) - 4.22 x 10^(-12)
x * 1.14 x 10^(-10) = 8.4 x 10^(-13)
x = (8.4 x 10^(-13)) / (1.14 x 10^(-10))
x = 0.0074 liters
Therefore, the student would need to prepare approximately 0.0074 liters or 7.4 mL of the hypochlorous acid to prepare the buffer solution.
To figure out the volume of hypochlorous acid (HOCl) that the student would need to prepare the buffer solution, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
In this equation, pH is the desired pH of the buffer, pKa is the acid dissociation constant of hypochlorous acid (which is determined by the temperature and can be found in a reference), [A-] is the concentration of the conjugate base (sodium hypochlorite, NaOCl), and [HA] is the concentration of the weak acid (hypochlorous acid, HOCl).
Since we are not given the pH and pKa values, we can assume that the desired pH of the buffer is the pKa value of hypochlorous acid (this provides maximum buffering capacity). The pKa of hypochlorous acid is approximately 7.53, so we will assume the desired pH is 7.53.
Now, rearrange the Henderson-Hasselbalch equation to solve for [A-]/[HA]:
[A-]/[HA] = 10^(pH - pKa)
Substitute the values into the equation:
[A-]/[HA] = 10^(7.53 - 7.53) = 10^0 = 1
This means that the ratio of [A-] to [HA] in the buffer solution is 1:1.
Since the concentration of the conjugate base is 0.046 M, we need to prepare a solution with an equal concentration of the weak acid. Therefore, the required concentration of the weak acid (hypochlorous acid) is also 0.046 M.
Now we can calculate the volume of hypochlorous acid that is needed using the equation:
Volume (in L) = moles / concentration
First, calculate the moles of hypochlorous acid required:
moles = concentration x volume
moles = 0.046 M x 0.250 L = 0.0115 mol
Then, calculate the volume of hypochlorous acid needed:
Volume (in L) = 0.0115 mol / 0.365 M = 0.0315 L
Therefore, the student would need to prepare 0.0315 L (31.5 mL) of hypochlorous acid to prepare the buffer solution.