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?
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
To draw the pH-titration curve for the acid-base reaction between 10 mL of 0.1M H3PO4 and 0.1 M NaOH, you need to plot the pH against the volume of the NaOH solution added.
Here's how you can construct the curve:
1. Start by plotting the initial pH. Since you have 10 mL of 0.1M H3PO4, the initial pH can be calculated using the formula pH = -log[H+]. For H3PO4, the initial pH can be determined by calculating the concentration of the hydrogen ions (H+). In this case, it would be -log(0.1) = 1.
2. From the initial pH, the first equivalence point can be determined. For a tri-protic acid like phosphoric acid (H3PO4), it has three stages of deprotonation. Each stage corresponds to a different equivalence point.
- The first equivalence point occurs when one hydrogen ion of H3PO4 is neutralized. Based on your given information, the logK value at this point is 2. So, at this point, you can plot a small jump in pH on your graph.
3. The second equivalence point occurs when the second hydrogen ion of H3PO4 is neutralized. Using the given logK value of 7, you can plot another jump in pH at this point.
4. Finally, the third equivalence point occurs when the last hydrogen ion of H3PO4 is neutralized. The logK value at this point is 12. Again, you can plot a jump in pH to represent this.
Note that between each of these equivalence points, you should expect gradual changes in pH, rather than sudden jumps.
5. Once you have plotted all the equivalence points, you can connect them with smooth curves to create the pH-titration curve. The pH should gradually increase as you add NaOH until each equivalence point, and then it will increase more rapidly after each equivalence point.
In summary, the x-axis represents the mL of NaOH added, and the y-axis represents the pH. The logK values indicate the pH values at the respective equivalence points, where you will observe small jumps on the graph.