pH = -log (H^+)
2.50 = -log (H^+)
(H^+) = 0.00316 M is current.
What you want is
3.10 = -log (H^+)
(H^+) = 0.00125 M
Plug in to the dilution formula.
mL1 x M1 = mL2 x M2
1000 mL x 0.00316 = mL2 x 0.00125
Solve for mL2 = total volume
Then assuming volumes are additive
total volume - 1000 mL = mL to be added.
The pH of a solution of HCL in water is found to be 2.50.
What volume of water would you add to 1.00L of this solution to raise the pH to 3.10?
1 answer
1 answer