You know pH, I would convert to pOH, then to (OH^-). That will be (OH^-) = ?? moles/L of solution. You take 10.00 mL of that and dilute to 250 mL; therefore, the new (OH^-) = ?? moles/L x 10/250 = xx moles/L= xx M of the solution to be titrated.
Then titration is
mL acid x M acid = mL base x M base
Check my thinking. I did NOT correct for the 2OH^- in Sr(OH)2 since we always worked directly with pH, pOH, etc., and never converted to molarity Sr(OH)2
The pH of saturated Sr(OH)2 (aq) is found to be 13.12. A 10.0 mL sample of saturated Sr(OH)2 (aq) is diluted to 250.0 mL in a volumetric flask. A 10.0 mL sample of the diluted Sr(OH)2 (aq) is transferred to a beaker, and some water is added. The resulting solution requires 26.4 mL of a HCl solution for its titration.
What is the molarity of this HCl solution?
2 answers
9.987*10^-4