posted by Brian .
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?