Asked by Finding the pH of a weak acid?
A vitamin C tablet containing 250 mg of ascorbic acid (C6H7O6; Ka = 8.0 x 10-5 is dissolved in a 250 mL glass of water. What is the pH of the solution?
i tried answering this question multiple times but i still get it wrong.... can someone help? step-by-step please.
what i did was....
make a chemical equation
C6H8O6 + H20 <--> C6H7O6- + H30+
but i fell like my equation is off....
but i did an ICE table and used it to find x then the pH
i tried answering this question multiple times but i still get it wrong.... can someone help? step-by-step please.
what i did was....
make a chemical equation
C6H8O6 + H20 <--> C6H7O6- + H30+
but i fell like my equation is off....
but i did an ICE table and used it to find x then the pH
Answers
Answered by
DrBob222
HC is ascorbic acid with Ka = 8.0E-5.
moles HC = 0.250g/molar mass HC = ?.
M = moles/0.250L = ? which I will call y.
............HC ==> H^+ + C^-
initial.....yM......0.....0
change.......-x......x....x
equil.......y-x......x....x
Ka = (H^+)(C^-)/(HC)
Substitute and solve for x = (H^+) and convert to pH.
This looks like a low molarity; you MAY need to solve the quadratic.
moles HC = 0.250g/molar mass HC = ?.
M = moles/0.250L = ? which I will call y.
............HC ==> H^+ + C^-
initial.....yM......0.....0
change.......-x......x....x
equil.......y-x......x....x
Ka = (H^+)(C^-)/(HC)
Substitute and solve for x = (H^+) and convert to pH.
This looks like a low molarity; you MAY need to solve the quadratic.
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