how much 10M HNO3 must be added to 1L of a buffer that is .010M acetic acid and .10M sodium acetate to reduce the pH to 5. pKa=4.75

millimoles acetic acid = 1000 mL x 0.01M = 10

mmole sodium acetate = 1000 mL x 0.1M = 100.

5.00 = 4.74 + log(base/acid)
base = 100-x
acid = 10+x
Solve for x where x = mols HNO3 added.
Then convert to volume needed of the 10M HNO3.

To determine how much 10M HNO3 must be added to reduce the pH of the buffer solution to 5, we need to make use of the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the pKa value and the concentrations of the acid and conjugate base.

The Henderson-Hasselbalch equation is given by:

pH = pKa + log([A-]/[HA])

Where:
pH = desired pH (5 in this case)
pKa = acid dissociation constant (-log Ka) = 4.75 (given)
[A-] = concentration of the conjugate base (sodium acetate) = 0.10 M
[HA] = concentration of the acid (acetic acid) = 0.010 M

Now, we can rearrange the equation to solve for the concentration ratio [A-]/[HA]:

[A-]/[HA] = 10^(pH - pKa)

Substituting the given values:

[A-]/[HA] = 10^(5 - 4.75)
[A-]/[HA] = 10^0.25
[A-]/[HA] = 1.78

This means that the concentration of the conjugate base ([A-]) in the buffer solution should be approximately 1.78 times higher than the concentration of the acid ([HA]) to achieve a pH of 5.

Since the initial volume of the buffer is 1L, the moles of acetic acid and sodium acetate are the same as their concentrations. Therefore, the moles of sodium acetate needed can be calculated as:

moles of sodium acetate = (1L) * (0.10 mol/L) = 0.10 mol

To determine the moles of acetic acid required, we multiply the moles of sodium acetate by the concentration ratio:

moles of acetic acid = (0.10 mol) / (1.78) = 0.056 mol

Finally, we can calculate the volume of 10M HNO3 required:

volume of 10M HNO3 = (moles of acetic acid) / (10M)
volume of 10M HNO3 = (0.056 mol) / (10 mol/L)
volume of 10M HNO3 = 0.0056 L or 5.6 mL

Therefore, you would need to add approximately 5.6 mL of 10M HNO3 to 1L of the buffer solution to reduce the pH to 5.