How much 10.0 M HNO3 must be added to 1.00 L of a buffer that is 0.0100 M acetic acid and 0.100 M sodium acetate to reduce the pH to 5.00? pKa of acetic acid= 4.75
1000 mL x 0.01M = 10 millimols HAc
1000 mL x 0.1M = 100 mmols NaAc
..........Ac^- + H^+ ==> HAc
I........100.....0........10
add HNO3.........x...........
C........-x.....-x.........+x
C.......100-x....0........10+x
pH = pKa + log [(100-x)/(10+x)]
Solve for x = mmols HNO3 needed.
M = mmols/mL. You know mmols and M, solve for mL HNO3. I think the answer is about 3 mL.
To find out how much 10.0 M HNO3 must be added to the buffer solution, we can use the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
In this equation, pH is the desired pH (5.00 in this case), pKa is the dissociation constant of the weak acid (4.75 in this case), [A-] is the concentration of the conjugate base (sodium acetate), and [HA] is the concentration of the weak acid (acetic acid).
Since we want to reduce the pH, the [A-]/[HA] ratio must be increased. Adding a strong acid, such as HNO3, will increase the concentration of [A-] as it reacts with sodium acetate.
Let's calculate the current [A-]/[HA] ratio in the buffer solution:
[A-]/[HA] = [sodium acetate]/[acetic acid]
= 0.100 M / 0.0100 M
= 10
Now let's calculate the ratio required to achieve a pH of 5.00 using the Henderson-Hasselbalch equation:
5.00 = 4.75 + log([A-]/[HA])
Rearranging the equation:
log([A-]/[HA]) = 5.00 - 4.75
= 0.25
Taking the antilog of both sides:
[A-]/[HA] = 10^0.25
≈ 1.778
Now we need to increase the [A-]/[HA] ratio from 10 to 1.778. This change requires us to add 10 mL of HNO3 per 1.00 L of buffer solution.
To find the amount of 10.0 M HNO3 needed:
Amount of HNO3 = Volume of HNO3 × Concentration of HNO3
= 10 mL × 10.0 M
= 100 mL
= 0.100 L
Therefore, you would need to add 0.100 L (or 100 mL) of 10.0 M HNO3 to 1.00 L of the given buffer solution in order to reduce the pH to 5.00.
To reduce the pH of the buffer to 5.00, we need to calculate how much 10.0 M HNO3 needs to be added.
First, we need to find the initial concentrations of acetic acid (CH3COOH) and acetate ions (CH3COO-) in the buffer solution.
Given:
Initial volume of the buffer = 1.00 L
Initial concentration of acetic acid (CH3COOH) = 0.0100 M
Initial concentration of sodium acetate (CH3COONa) = 0.100 M
Since acetate ions come from sodium acetate, the initial concentration of acetate ions is also 0.100 M.
Next, we need to consider the Henderson-Hasselbalch equation to relate the pH to the concentrations of acetic acid and acetate ions:
pH = pKa + log([CH3COO-]/[CH3COOH])
In this equation, pKa is the negative logarithm of the acid dissociation constant, which is given as 4.75.
Rearranging the equation, we have:
[CH3COO-]/[CH3COOH] = 10^(pH - pKa)
Substituting the given values, we find:
[CH3COO-]/[CH3COOH] = 10^(5.00 - 4.75)
[CH3COO-]/[CH3COOH] = 10^(0.25)
[CH3COO-]/[CH3COOH] ≈ 1.78
Now, we can determine the moles of acetic acid and acetate ions using the molarity and volume of the buffer solution:
Moles of CH3COOH = 0.0100 M x 1.00 L = 0.0100 mol
Moles of CH3COO- = 0.100 M x 1.00 L = 0.100 mol
To maintain the ratio of acetate ions to acetic acid at 1.78 (as calculated above), we need to add enough HNO3 to neutralize the excess acetate ions and convert them into acetic acid.
Considering the balanced equation:
CH3COO- + HNO3 → CH3COOH + NO3-
We can see that 1 mole of HNO3 will react with 1 mole of CH3COO-.
Therefore, the moles of HNO3 required to neutralize the excess acetate ions is:
Moles of HNO3 = Moles of CH3COO- - Moles of CH3COOH
Moles of HNO3 = 0.100 mol - 0.0100 mol
Moles of HNO3 = 0.0900 mol
Finally, we can calculate the volume of 10.0 M HNO3 required using the relation between moles, concentration, and volume:
Volume of HNO3 = Moles of HNO3 / Concentration of HNO3
Volume of HNO3 = 0.0900 mol / 10.0 M
Volume of HNO3 = 0.00900 L or 9.00 mL
Therefore, you would need to add 9.00 mL of 10.0 M HNO3 to reduce the pH of the buffer to 5.00.