I really struggled with understanding titrations until I worked my way through this theoretical titration of carbonic acid with NaOH, so I figured I’d post it here in case anyone else needed help! This can also apply to titration of amino acids!
Here are the key ideas to know:
• Essentially, the number of protons you have to neutralize = the number of half-equivalence points = the number of full equivalence points
• The full equivalence point is where you’ve COMPLETELY removed a proton and you have 100% the resulting deprotonated conjugate base
• The half-equivalence point is the same thing as the buffering region! Thus, at a half equivalence point, you have a 50:50 mixture of an acid and its deprotonated conjugate base!
• M1n1V1 = M2n2V2 (memorize this equation like your life DEPENDS on it lmao); this equation will tell you the conditions at the equivalence point of the titration, which is when you’ve completely neutralized all your acid!
– M = molarity
– n = the number of “equivalents” (e.g., for NaOH, n = 1 because it gives 1 equivalent of base; for H2SO4, n = 2 because it gives 2 equivalents of acidic protons)
– V = volume (in mL or L)
☆ Say you’re performing the titration of 50 mL 0.1 M H2CO3 with 0.2 M NaOH
• Use your formula; you should get that you require 50 mL of NaOH to completely neutralize all your H2CO3. 50 mL NaOH is equivalent to two equivalents of NaOH.
• What does “completely neutralized” H2CO3 look like? At the equivalence point, you’ve stripped off BOTH of H2CO3’s protons, leaving you with CO3(2-)!
• Since you know 50 mL NaOH strips off BOTH protons, you can thus infer that 25 mL NaOH only strips off ONE proton, giving you HCO3-!
With this in mind, let’s work through the titration:
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At very acidic pH values, H2CO3 exists predominantly at 100% H2CO3
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When you have added 0.5 equivalents of NaOH (12.5 mL), H2CO3 exists as a 50:50 mixture of HCO3- and H2CO3! This is the FIRST half-equivalence point (buffering region)! Remember what a buffer is? A buffer is a 50:50 mixture of either an acid and a salt containing its conjugate base (e.g., acetic acid and sodium acetate) or a 50:50 mixture of a base and a salt containing its conjugate acid (e.g., ammonia and ammonium chloride)! This is exactly what you have at the half-equivalence point: a buffer! “50:50” = “half and half” = “half-equivalence point”! The half-equivalence point/buffering region would appear as a flat line on a titration curve
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When you have added 1 equivalent of NaOH (25 mL), H2CO3 exists as 100% HCO3-! This is the FIRST full equivalence point, signalling that you’ve COMPLETELY removed the first proton from H2CO3 and now ONLY have HCO3-! Again, there is no 50:50 mixture, so this is a FULL equivalence point, NOT a half-equivalence point! This would appear as a sharp vertical line on a titration curve!
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When you have added 1.5 equivalents of NaOH (37.5 mL), you will have a 50:50 mixture of HCO3- and CO3(2-)! This is the SECOND half-equivalence point/buffering region! Again, it appears as a flat line!
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When you’ve added 2 equivalents of NaOH (50 mL), you will have a 100% mixture of CO3(2-)! This is the SECOND and FINAL equivalence point! Again, it appears as a sharp vertical line!
And that’s it! As long as you understand the equation and the concept of equivalents, you should be golden! I really hope this helps out anyone who needs it. Good luck!
