You have solutions of 0.200 M HNO2 and 0.200 M KNO2 (Ka for HNO2 = 4.00 10-4). A buffer of pH 3.000 is needed. What volumes of HNO2 and KNO2 are required to make 1 liter of buffered solution?
How does the henderson hasslebache equation help find the volume?
pH = pKa + log [(base)/(acid)]
You know Ka. Calculate pKa.
NO2^- is the base.
HNO2 is the acid.
First, plug in pH = 3.00, the pH of the solution you want, and calculate the (base)/(acid) ratio.
That gives you (base) = factor x (acid).
Then you know mLbase*Mbase = mLacid*Macid AND you know mLacid + mL bse = 1000.
You solve for mLacid and mL base. If I didn't goof, it is something like 700 mL acid and 300 mL base.
Those volumes I quoted are not the exact volumes but they are close to the correct values.
The Henderson-Hasselbalch equation is an equation that relates the pH of a buffer solution to the ratio of the concentrations of its acidic and conjugate base components. It is given by:
pH = pKa + log([A-]/[HA])
Where:
- pH is the desired pH of the buffer solution.
- pKa is the negative logarithm (base 10) of the acid dissociation constant (Ka) of the acid component.
- [A-] is the concentration of the conjugate base.
- [HA] is the concentration of the acid.
By rearranging the Henderson-Hasselbalch equation, it is possible to solve for the ratio [A-]/[HA]:
[A-]/[HA] = 10^(pH - pKa)
Now, let's apply this equation to solve the given problem.
1. Firstly, identify the given data:
- pH = 3.000
- pKa = -log(Ka) = -log(4.00 x 10^(-4)) = 3.40
2. Substitute the values in the Henderson-Hasselbalch equation:
10^(3.000 - 3.400) = [A-]/[HA]
3. Calculate the value of [A-]/[HA]:
10^(-0.400) = [A-]/[HA]
[A-]/[HA] = 0.398
The ratio of [A-] to [HA] in the buffer solution should be approximately 0.398.
Now, let's proceed to find the volumes of HNO2 and KNO2 required to make 1 liter (1000 mL) of buffered solution:
4. Assume the total volume of the buffer solution is 1000 mL.
Let the volume of HNO2 be x mL, and the volume of KNO2 be (1000 - x) mL.
5. Calculate the moles of HNO2 and KNO2 required to achieve the desired concentration in the buffer solution:
Moles of HNO2 = (0.200 M) * (x mL/1000 mL)
Moles of KNO2 = (0.200 M) * ((1000 - x) mL/1000 mL)
6. Use the molar ratios to determine the ratio of the volumes of HNO2 to KNO2:
(Moles of HNO2)/(Moles of KNO2) = [HNO2]/[KNO2] = [HA]/[A-] = 1/0.398
(0.200 M) * (x mL/1000 mL) / (0.200 M) * ((1000 - x) mL/1000 mL) = 1 / 0.398
x / (1000 - x) = 1 / 0.398
7. Further rearrange and solve the equation for x (the volume of HNO2):
0.398 * x = 1000 - x
0.398 * x + x = 1000
1.398 * x = 1000
x = 1000 / 1.398
x ≈ 715.25 mL
So, approximately 715.25 mL of HNO2 and (1000 - 715.25) mL ≈ 284.75 mL of KNO2 are required to make 1 liter of the buffered solution.
To find the volumes of HNO2 and KNO2 required to make 1 liter of buffered solution, we can use the Henderson-Hasselbalch equation. This equation relates the pH of a buffer solution to the concentrations of the acidic and basic components of the buffer.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
Where:
pH = desired pH of the buffer solution
pKa = -log(Ka) (Ka is the acid dissociation constant)
[A-] = concentration of the conjugate base (in this case, KNO2)
[HA] = concentration of the acid (in this case, HNO2)
In this problem, we are given the desired pH of the buffer solution (pH = 3.000) and the concentration of HNO2 (0.200 M). We are also given the Ka for HNO2 (4.00 x 10^-4).
To find the concentration of KNO2, we can rearrange the Henderson-Hasselbalch equation and solve for [A-]:
[A-] = 10^(pH - pKa) * [HA]
In this case, [HA] is the concentration of HNO2, which is 0.200 M. Plugging in the values:
[A-] = 10^(3 - (-log(4.00 x 10^-4))) * 0.200 M
= 0.630 M
Now that we have the concentrations of HNO2 (0.200 M) and KNO2 (0.630 M), we can calculate the volumes needed to make 1 liter of the buffered solution.
Let V1 be the volume of HNO2 and V2 be the volume of KNO2. Since the total volume is 1 liter, we have:
V1 + V2 = 1 liter
Now, we need to convert the concentrations to moles:
Moles of HNO2 = Molarity * Volume
= 0.200 M * V1
Moles of KNO2 = Molarity * Volume
= 0.630 M * V2
Since moles = volume * concentration, we can rewrite the above expressions as:
V1 = Moles of HNO2 / 0.200
= Moles of HNO2 / (0.200 * 1 liter)
V2 = Moles of KNO2 / 0.630
= Moles of KNO2 / (0.630 * 1 liter)
Since we want to make a 1 liter buffer solution:
V1 + V2 = 1 liter
Now, we can substitute the expressions for V1 and V2:
(Moles of HNO2 / (0.200 * 1 L)) + (Moles of KNO2 / (0.630 * 1 L)) = 1 L
Simplifying the equation, we get:
Moles of HNO2 / 0.200 + Moles of KNO2 / 0.630 = 1
Now that we have the equation in terms of moles, we need to calculate the moles of HNO2 and KNO2 using the concentrations and volumes provided. Finally, we can substitute those values into the equation to find the volumes of HNO2 and KNO2 needed to make 1 liter of the buffered solution.