Using a 0.25 M phosphate buffer with a pH of 6.6, you add 0.71 mL of 0.51 M HCl to 49 mL of the buffer. What is the new pH of the solution? (Enter your answer to three significant figures.)
Using a 0.25 M phosphate buffer with a pH of 6.6, you add 0.71 mL of 0.51 M NaOH to 49 mL of the buffer. What is the new pH of the solution? (Enter your answer to three significant figures.)
I just don't understand how to do this and what the answers are.
You use the Henderson-Hasselbalch equation. Each of these problems has two parts to it. First, you must use the HH equation to determine the (acid) and (base). Second, you plug in the added (H^+) from the HCl and calculate the new pH from the HH equation for problem 1 OR plug in the added (OH^-) from the NaOH and calculate the new pH from the HH equation for problem 2. Here is the outline for problem 1.
pH = pK2 + (B)/(A) where (B) = (base) and (A)-= (acid)
6.6 = 7.20 + B/A. You should look up pK2 for H3PO4 in your text and use that. The 7.20 is from my references.
Solve for B/A. That's two unknowns so you need another equation. You get that from the problem. It says that (A) + (B) = 0.25 M. Solve these two equations simultaneously for (A) and (B). That part isn't chemistry, just math. Post your work if you get stuck. For your reference, I obtained about 0.2M for acid and 0.05M for base. Then you go to part two for the first problem.
I prefer to work in millimoles from here. You have
millimoles buffer = M x mL = 0.25M x 49 mL = 12.25
millimoles base = M x mL = 0.05 x 49 mL = 2.45 = (HPO4^-2)
millimoles acid = M x mL -= 0.20 x 49 mL = 9.8 = (H2PO4^-)
millimoles HCl added = M x mL = 0.51M x 0.71 mL = 0.36
Set up an ICE chart as follows:
............HPO4^2- + H^+ = H2PO4^-
I...........2.45...........0.36........9.80
add .......................0.36..................
C........-0.36...........-0.36........+0.36
E.........2.09................0.........1.02
Take the E values in millimoles, convert to concentrations (M = millimoles/mL) and plug into the HH equation to calculate the new pH of the solution.
Follow the same steps for the added NaOH.
Post all of your work if you get stuck.
Where are you getting the 1.02 from for the equilibrium
To calculate the new pH of the solution after adding an acid (HCl) or a base (NaOH) to a buffer, you need to consider the Henderson-Hasselbalch equation.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log ([A-]/[HA])
Where:
pH = the pH of the solution
pKa = the acid dissociation constant of the buffer
[A-] = the concentration of the conjugate base
[HA] = the concentration of the weak acid
For the first scenario, where you add HCl to the buffer:
1. Calculate the new concentration of the acid and the conjugate base:
Starting concentration of the acid (HA): 0.25 M x 49 mL = 12.25 mmol
Concentration of the acid after addition (HA): 12.25 mmol - (0.71 mL x 0.51 M) = 11.86 mmol
Concentration of the conjugate base (A-): 0.71 mL x 0.51 M = 0.362 mmol
2. Calculate the new pH using the Henderson-Hasselbalch equation:
pKa (phosphate buffer) is approximately 7.21 (at 25°C)
pH = 7.21 + log (0.362 mmol / 11.86 mmol)
pH ≈ 7.21 + log (0.0306)
pH ≈ 7.21 + (-1.515)
pH ≈ 5.70
So, the new pH of the solution after adding HCl is approximately 5.70.
For the second scenario, where you add NaOH to the buffer:
1. Calculate the new concentration of the base and the conjugate acid:
Starting concentration of the conjugate acid (HA): 0.25 M x 49 mL = 12.25 mmol
Concentration of the conjugate acid after addition (HA): 12.25 mmol - (0.71 mL x 0.51 M) = 11.86 mmol
Concentration of the base (A-): 0.71 mL x 0.51 M = 0.362 mmol
2. Calculate the new pH using the Henderson-Hasselbalch equation:
pKa (phosphate buffer) is still approximately 7.21 (at 25°C)
pH = 7.21 + log (0.362 mmol / 11.86 mmol)
pH ≈ 7.21 + log (0.0306)
pH ≈ 7.21 + (-1.515)
pH ≈ 5.70
So, the new pH of the solution after adding NaOH is approximately 5.70.
Therefore, in both scenarios, the new pH of the solution is approximately 5.70.
To answer these questions, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the concentration of its acid and conjugate base components.
The Henderson-Hasselbalch equation is given by pH = pKa + log ([A-]/[HA]), where [A-] is the concentration of the conjugate base and [HA] is the concentration of the acid.
In the first scenario, we are adding HCl, which is a strong acid, to the buffer solution. Since it is a strong acid, it will completely dissociate into its ions. The concentration of H+ ions will increase, and as a result, the pH of the solution will decrease.
We can calculate the concentration of H+ ions after adding HCl by using the formula: [H+] = (moles of HCl) / (total volume of solution).
In this case, we are adding 0.71 mL of 0.51 M HCl to a total volume of 49 mL. Let's calculate the moles of HCl added:
moles of HCl = (0.71 mL) * (0.51 mol/L) * (1 L / 1000 mL) = 0.00036321 mol
Next, let's calculate the new concentration of H+ ions:
[H+] = (0.00036321 mol) / (0.04971 L) = 0.007312 M
Now that we have the concentration of H+ ions, we can use the Henderson-Hasselbalch equation to calculate the new pH. Given that the original pH is 6.6 and the pKa of the phosphate buffer is around 7.2, we can substitute the values into the equation:
pH = 7.2 + log ([A-]/[HA])
pH = 7.2 + log (0.25/0.25) // Since the buffer has equal concentrations of acid and conjugate base, [A-]/[HA] = 1
pH = 7.2 + log (1)
pH = 7.2
So, the new pH of the solution after adding HCl is 7.2 (to three significant figures).
In the second scenario, we are adding NaOH, which is a strong base, to the buffer solution. Similarly, the concentration of OH- ions will increase, and the pH of the solution will increase.
Using the same calculations as before:
The moles of NaOH added = (0.71 mL) * (0.51 mol/L) * (1 L / 1000 mL) = 0.00036321 mol.
The new concentration of OH- ions = (0.00036321 mol) / (0.04971 L) = 0.007312 M.
Now, to calculate the new pH, we can use the Henderson-Hasselbalch equation again:
pH = 7.2 + log ([A-]/[HA])
pH = 7.2 + log (0.25/0.25)
pH = 7.2 + log (1)
pH = 7.2
So, the new pH of the solution after adding NaOH is also 7.2 (to three significant figures).