Titration of a Weak Base with a Strong Acid
- Page ID
- 221878
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At this point, there is no BH+ or OH- in the analyte solution—the molarities of these species are zero, as observed in the ICE table. Because the stopcock has not yet been released, there is no acid in the flask to react with the base and yield products BH+ and OH-. Once the strong acid is released into the flask, however, the BH+ and OH- begin to form. Reaction of weak base and strong acid.
H+(aq) + ClO-(aq) -> HClO(aq)
- Start with 100 ml(1.00 M) = 0.100 moles of ClO-.
- If you add 50ml(1.00 M) = 0.05 moles of HCl to the base, the reaction in the previous step will consume all of the H+, leaving 0.100 - 0.050 = 0.050 moles of ClO-.
- Because we added 50 ml of acid to 100 ml of base, we have a solution volume of 150 ml.
- The concentration of ClO- after adding the base is [ClO-] = 0.050 moles/0.150 L = 0.333 M.
- 0.050 moles of HClO also forms, thus the concentration of HClO is also 0.333 M.
- The solution above is a buffer of the weak acid HClO and the conjugate base ClO-.
- Now we can solve for the pH of a buffer by using the Henderson-Hasselbalch approximation.
References
- Petrucci, Ralph H., et al. General Chemistry: Principles and Modern Applications.
- Upper Saddle River, NJ: Prentice Hall, 2007
http://cartage.org.lb/en/themes/Sciences/Chemistry/Inorganicchemistry/AcidsBases/Acidsbasesindex/weakbasetitration.htm http://www.chem.ubc.ca/courseware/pH/section14/content.html
Contributors
- Christine Chang, Kelly Cox