Asked by TORI
A student takes a 1.00 mL aliquout of 5.00*10^-4 M HCl solution and dilutes it to the mark with water in a 1000.0 mL volumetric flask.
(a)What is the expected pH for the final solution at 25C?
(b)What would be the expected pH if this solution were later used at a temperature of 20C?
(a)What is the expected pH for the final solution at 25C?
(b)What would be the expected pH if this solution were later used at a temperature of 20C?
Answers
Answered by
DrBob222
(a) I would approach a this way.
HOH ==> H^+ + OH^-
Kw = (H^+)(OH^-)
Look up Kw for H2O @ 25C (usually we use 1E-14; however, looking at the b part that may not be the proper one to use at 25C). As a first approximation, I would plug in 5E-7 for (H^+) and solve for OH^-, then OH^- + 5E-7 should come very close. If you want it closer, it will be
(5E-7+x)(x)=Kw and (H^+)=(5E-7+x)
b. Same procedure but use the new Kw for 20C and not 25C.
HOH ==> H^+ + OH^-
Kw = (H^+)(OH^-)
Look up Kw for H2O @ 25C (usually we use 1E-14; however, looking at the b part that may not be the proper one to use at 25C). As a first approximation, I would plug in 5E-7 for (H^+) and solve for OH^-, then OH^- + 5E-7 should come very close. If you want it closer, it will be
(5E-7+x)(x)=Kw and (H^+)=(5E-7+x)
b. Same procedure but use the new Kw for 20C and not 25C.
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