Asked by Eugene
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?
What volume of water would you add to 1.00L of this solution to raise the pH to 3.10?
Answers
Answered by
DrBob222
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.
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.
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