Asked by Duffy
Calculate the pH of the resultant solution when 40.00 mL of 0.250 M H2SO4 are mixed with 30.00 mL of 0.450 M KOH.
The answer says it is 1.032...I get 12.70 (and I subtract that from 14 I still only get 1.3)
Thanks.
The answer says it is 1.032...I get 12.70 (and I subtract that from 14 I still only get 1.3)
Thanks.
Answers
Answered by
DrBob222
I believe the answer is in error.
H2SO4 does have a k2 but with excess KOH k2 for H2SO4 won't make any difference.
mmols H2SO4 = mL x M = 40.00 x 0.250 = 10.0 mmols.
mmols KOH = 30.00 x 0.450 = 13.50 mmoles.
13.5 - 10.0 = 3.5 mmoles KOH in excess.
(OH^-) = 3.5 mmols/70 mL = 0.0500 M OH.
pOH = 1.30.
pH = 12.70
H2SO4 does have a k2 but with excess KOH k2 for H2SO4 won't make any difference.
mmols H2SO4 = mL x M = 40.00 x 0.250 = 10.0 mmols.
mmols KOH = 30.00 x 0.450 = 13.50 mmoles.
13.5 - 10.0 = 3.5 mmoles KOH in excess.
(OH^-) = 3.5 mmols/70 mL = 0.0500 M OH.
pOH = 1.30.
pH = 12.70
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