What volume of 0.01 M NaOH should be required to raise the pH of a litre of 25 mM H2SO4 to 4.0?

I started out by attempting to find the initial pH of H2SO4 and determined I need 2.57 units of pH to raise it to 4. I know that somewhere I use the HH equation, but don't know what concentrations to use for [B-] or [BA], or how many mols of NaOH to add etc. I know to use that value to relate back to the initial conc. of NaOH, but don't know what to do there either.

Please help.

To solve this problem, we can calculate the amount of NaOH required using the Henderson-Hasselbalch equation and stoichiometry. Here's how you can proceed:

Step 1: Calculate the initial pH of the H2SO4 solution.
Given that the H2SO4 concentration is 25 mM, we can convert it to moles per liter (M) by dividing by 1000:
[H2SO4] = 25 mM / 1000 = 0.025 M

Since H2SO4 is a strong acid, it fully ionizes in solution. Therefore, the concentration of H+ is equal to the concentration of H2SO4:
[H+] = 0.025 M

To find the initial pH, you can use the equation:
pH = -log[H+]

Step 2: Calculate the amount of NaOH required to raise the pH to 4.
Given that you need to raise the pH from its initial value (calculated in step 1) to 4, you know the change in pH is 4 - initial pH.

Let's call the change in pH "ΔpH":
ΔpH = 4 - initial pH = 4 - [-log[H+]]

Step 3: Use the Henderson-Hasselbalch equation to relate the change in pH to the ratio of [B-] to [BA].
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([B-]/[BA])

In this context, NaOH (sodium hydroxide) acts as the base (B-) and H2SO4 acts as the acid (BA).

Since we're assuming the final pH will be 4, we need to find the corresponding pKa value. The pKa for H2SO4 is the negative logarithm of its Ka value.

- log(Ka) = pKa

Step 4: Calculate the ratio [B-]/[BA].
Rearrange the Henderson-Hasselbalch equation:
log([B-]/[BA]) = pH - pKa

Now, take the antilog of both sides to eliminate the logarithm:
[B-]/[BA] = 10^(pH - pKa)

Step 5: Calculate the amount of NaOH required.
The ratio of [B-]/[BA] is equal to the ratio of moles of NaOH added to moles of H2SO4 neutralized.

Since the concentration of NaOH is given as 0.01 M, the concentration of Na+ is also 0.01 M after complete ionization.

Let's assume you add x moles of NaOH to neutralize the H2SO4. As H2SO4 is a diprotic acid, it forms two moles of H+ ions for every mole of H2SO4, so it requires two moles of NaOH per mole of H2SO4 neutralized.

Therefore, the moles of Na+ added is 2x.

Now, we can set up the equation using the concentration and moles:
0.01 M (Na+) = (2x) / (Volume NaOH)

Where Volume NaOH is the total volume of NaOH needed to reach the desired pH.

Step 6: Solve for Volume NaOH.
Rearranging the equation:
Volume NaOH = (2x) / 0.01 M

To find x (the moles of NaOH required), we can use the stoichiometric ratio between NaOH and H2SO4. Since two moles of NaOH are required per mole of H2SO4, we can set up the equation:
x = (0.025 M)(Volume H2SO4)

where Volume H2SO4 is the total volume of H2SO4 in liters.

Step 7: Substitute x into the equation for Volume NaOH.
Volume NaOH = (2(0.025 M)(Volume H2SO4)) / 0.01 M

Simplifying the equation will give you the volume of NaOH needed in the correct units to raise the pH to 4.