Asked by Grace
Substance X (MW=356gm) is a weak acid. Its Ka is 1.7x10^-3. Calculate pH of725ml of the following solutions:
a)Containing 0.187 mol of substance x
b)containing 127gm of Substance x
I have been trying to figure this problem out for hours now and do not get it. Please help me and be very detailed. Thank you.
a)Containing 0.187 mol of substance x
b)containing 127gm of Substance x
I have been trying to figure this problem out for hours now and do not get it. Please help me and be very detailed. Thank you.
Answers
Answered by
DrBob222
Call the weak acid HA.
(HA) = mols/L = 0.187/0.725 = about 0.26M but you should be more accurate than that.
.......HA --> H^+ + A^-
I.....0.26M....0.....0
C.....-x.......x.....x
E....-0.26-x...x.....x
Ka = 1.7E-3 = (H^+)(A^-)/(HA)
Substitute into Ka expression and solve for x = (H^+) then convert to pH.
Part B is done the same way EXCEPT you don't have mols to start. You convert 127g to mols by mols = grams/molar mass, then M = mols/L solution and from there do the same process as above.
Post your work if you get stuck.
(HA) = mols/L = 0.187/0.725 = about 0.26M but you should be more accurate than that.
.......HA --> H^+ + A^-
I.....0.26M....0.....0
C.....-x.......x.....x
E....-0.26-x...x.....x
Ka = 1.7E-3 = (H^+)(A^-)/(HA)
Substitute into Ka expression and solve for x = (H^+) then convert to pH.
Part B is done the same way EXCEPT you don't have mols to start. You convert 127g to mols by mols = grams/molar mass, then M = mols/L solution and from there do the same process as above.
Post your work if you get stuck.
Answered by
Grace
When I solve for H^+ I put 1.7 x10^-3 x 0.26. I got 4.42 x 10^-4. My pH is 3.35. Is that right?
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