Asked by Sean
A 20.0-mL sample of 0.20M sodium acetate is titrated with 0.11M HCl(aq). What is the pH after the addition of 50.0-mL HCl(aq)? (Kb of CH3CO2 = 5.6x10^-10)
Answers
Answered by
DrBob222
acetic acid is HAc.
acetate ion is Ac^-
Ac^- + H^+ ==> HAc.
I would approach the problem this way.
You started with 0.004 moles (0.020L x 0.2M)Ac^- and added 0.050 L x 0.11 M HCl (0.0055 moles) You should have at the end 0.004 moles HAc plus 0.0015 moles H^+ in excess (total volume now 70 mL) from the strong acid HCl. So the pH should be largely determined by the strong acid. Its concn will be 0.0015 moles/0.070 L and you can work out the concn and pH from that. Will the HAc contribute to the overall pH? Some but I don't think very much. You can, however, make a quick calculation of the contribution from the HAc and I think it will be negligible. Check my thinking.
acetate ion is Ac^-
Ac^- + H^+ ==> HAc.
I would approach the problem this way.
You started with 0.004 moles (0.020L x 0.2M)Ac^- and added 0.050 L x 0.11 M HCl (0.0055 moles) You should have at the end 0.004 moles HAc plus 0.0015 moles H^+ in excess (total volume now 70 mL) from the strong acid HCl. So the pH should be largely determined by the strong acid. Its concn will be 0.0015 moles/0.070 L and you can work out the concn and pH from that. Will the HAc contribute to the overall pH? Some but I don't think very much. You can, however, make a quick calculation of the contribution from the HAc and I think it will be negligible. Check my thinking.
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