Yes, 2 is the first equivalence point
See this graph. Note the first equivalence point is HUGE compared to the third.https://www.google.com/search?q=image+H3PO4+and+NaOH&tbm=isch&tbo=u&source=univ&sa=X&ei=HCG3VJTBFJGTNuGXguAI&ved=0CCIQsAQ&biw=1024&bih=609#imgdii=_&imgrc=NiLLxKTNMnpqUM%253A%3Bnk13b6oui93J0M%3Bhttp%253A%252F%252Fchemwiki.ucdavis.edu%252F%2540api%252Fdeki%252Ffiles%252F26796%252F9e679e00ac436bc1b485369c9db8190f.jpg%253Fsize%253Dbestfit%2526width%253D600%2526height%253D416%2526revision%253D1%3Bhttp%253A%252F%252Fchemwiki.ucdavis.edu%252FWikitexts%252FUC_Davis%252FUCD_Chem_2B%252FUCD_Chem_2B%25253A_Larsen%252FUnit_III%25253A_Acids_and_Bases%252F16.5_Acid%2525E2%252580%252593Base_Titrations%3B1715%3B1191
Draw the pH-titration curve for the following, label x and y axes accordingly.
10 mL of 0.1M H3PO4 with 0.1 M NaOH (Logk values = 2, 7, 12)
I understand that the x-axis is mL of H3PO4 added and y-axis is the pH. But the logK values, is that the equiv-points for the tri-protic acid i.e at pH 2, H3PO4 loses H+ and becomes H2PO4 and thus on the graph it is a small jump?
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