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.