Asked by Mya
What is the pH when 25 mL of .20 M CH3COOH has been titrated with 35 mL of .10 M NaOH?
I got the answer .0375 but that is incorrect if you could show me steps maybe i can figure out what i did wrong thanks
I got the answer .0375 but that is incorrect if you could show me steps maybe i can figure out what i did wrong thanks
Answers
Answered by
DrBob222
It would have been easier if you had shown your work and let us tell you what you did wrong. CH3COOH = HAc
HAc + NaOH ==> NaAc + H2O
mmoles HAc = 25 x 0.20 M = 5.00
mmoles NaOH = 35 x 0.1 = 3.50
mmoles H2O formed = mmoles NaAc formed = 3.50.
mmoles NaOH left over = 0
mmoles HAc left over = 1.50
(HAc) = 1.5mmoles/60 mL = ??
(NaAc) = 3.5 mmoles/60 mL = ??
(H^+)(Ac^-)/(HAc) = Ka.
Substitute for Ac and HAc and Ka and solve for H^+, then convert to pH. I obtained approximately 5 but you need to do it more accurately.
HAc + NaOH ==> NaAc + H2O
mmoles HAc = 25 x 0.20 M = 5.00
mmoles NaOH = 35 x 0.1 = 3.50
mmoles H2O formed = mmoles NaAc formed = 3.50.
mmoles NaOH left over = 0
mmoles HAc left over = 1.50
(HAc) = 1.5mmoles/60 mL = ??
(NaAc) = 3.5 mmoles/60 mL = ??
(H^+)(Ac^-)/(HAc) = Ka.
Substitute for Ac and HAc and Ka and solve for H^+, then convert to pH. I obtained approximately 5 but you need to do it more accurately.
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