Consider 57.5 mL of a solution of weak acid HA (Ka = 1.00 10-6), which has a pH of 3.800. What volume of water must be added to make the pH = 6.000?

Change pH to (H^+).

Write Ka expression.
Plug in (H^+) and (A^-) and solve for (HA).
Then change pH 6.00 to (H^+). Write Ka expression, plug in (H^+), (A^-) and solve for new (HA). Then calculate how much water must be added to convert concn from 3.80 material to concn you want at 6.00 concn.

how do u calc how much water is needed?

Molarity*liters of 1 soln = molarity*liters of other solution.

for the second set i put Ph=6 and got H+=1e-6 and then HA=1e6

i think that's wrong.. whats wrong with it?

To solve this problem, we need to use the Henderson-Hasselbalch equation, which relates the pH of a solution to the pKa of the weak acid and the concentrations of the acid and its conjugate base.

The Henderson-Hasselbalch equation is given by:

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

Where:
pH = the pH of the solution
pKa = the negative logarithm of the acid dissociation constant (Ka)
[A-] = the concentration of the conjugate base
[HA] = the concentration of the weak acid

Since the pH is given as 3.800, we can substitute the values into the equation:

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

We know that pKa = -log(Ka) = -log(1.00 × 10^(-6)) = 6.

So, the equation becomes:

3.800 = 6 + log([A-]/[HA])

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

log([A-]/[HA]) = 3.800 - 6

log([A-]/[HA]) = -2.200

Next, to find the concentration ratio, we need to convert the logarithmic equation back into exponential form:

[A-]/[HA] = 10^(-2.200)

[A-]/[HA] = 0.006309

Now that we have the concentration ratio of [A-]/[HA] = 0.006309, we can use it to solve the problem.

Using the formula for the pH of a weak acid solution:

pH = -log([HA])

We can plug in the given pH (6.000) and solve for the concentration of HA:

6.000 = -log([HA])

Taking the antilog of both sides:

[HA] = 10^(-6.000)

[HA] = 0.001

Now, we need to find the volume of water that must be added to make the pH = 6.000. The concentration of the weak acid will remain constant, so the total volume will be:

Total volume = Initial volume of solution + Volume of water added

Since the initial volume is given as 57.5 mL and we are adding water, the final volume will simply be:

Final volume = 57.5 mL + Volume of water added

We need to find the volume of water added, so we subtract the initial volume from the final volume:

Volume of water added = Final volume - Initial volume
Volume of water added = (57.5 mL + Volume of water added) - 57.5 mL
Volume of water added = Volume of water added

This means that any volume of water added will result in the same final volume. Therefore, any volume of water can be added to achieve a pH of 6.000.