Asked by tomi
The pH of saturated Sr(OH)2 is found to be 13.12. A 15.0ml sample of saturated Sr(OH)2 is diluted to 250.0ml in a volumetric flask. A 15.0ml sample of the diluted Sr(OH)2 is transferred to a beaker, and some water is added. The resulting solution requires 29.3ml of a HCl solution for its titration.
What is the molarity of this HCL solution?
What is the molarity of this HCL solution?
Answers
Answered by
DrBob222
The pH of the satd soln is 13.12; therefore, the pOH = 14 - 13.12 = 0.88
From this and pOH = -log(OH^-) you can find (OH^-) = about 0.12 (you need to do it exactly).
This was diluted from 15 mL to 250 so the (OH^-) in the 250 mL flask is 0.12 x 15/250 = about 0.0072 M (again you need to go through it to get a more exact answer).
So now you have a 15.0 mL sample of 0.0072 M OH^- titrated with 29.3 mL of ?M HCl.
mLbase x M base = mLacid x M acid.
Solve for M acid.
From this and pOH = -log(OH^-) you can find (OH^-) = about 0.12 (you need to do it exactly).
This was diluted from 15 mL to 250 so the (OH^-) in the 250 mL flask is 0.12 x 15/250 = about 0.0072 M (again you need to go through it to get a more exact answer).
So now you have a 15.0 mL sample of 0.0072 M OH^- titrated with 29.3 mL of ?M HCl.
mLbase x M base = mLacid x M acid.
Solve for M acid.
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