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alphabeta

  • one year ago

Titration question (attached file). Can someone tell me, in as much detail as possible, how I should approach this exercise?

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  1. alphabeta
    • one year ago
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  2. CherokeeGutierrez
    • one year ago
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    idk if this would help but your bascially copying something on the top graph to the bottom.

  3. welshfella
    • one year ago
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    yes with the difference that you include the pH range of the bromophenol blue indicator

  4. alphabeta
    • one year ago
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    Well, I don't believe that the pH range of the indicator is the only difference? For example, the initial pH of the acid should be different I guess (but how do I know where? Is it a value that I just have to learn by heart, or?). Also the half-equivalence point should be different, since you have a weak acid with strong base in this case compared to the first graph (but again, how do I figure this out?).

  5. cuanchi
    • one year ago
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    OK dear: you are in the right way, the propanoic acid is a weak acid with a pKa 4.88, and the potassium hydroxide is a strong base. Then the pHrange of the indicator is 3.0-4.6. The pH of the acid at 0mL titration can be calculated with the formula \[[H ^{+}]= \sqrt{Ka \times Ca}\] where pKa= -log Ka = 4.88 Ka= 10^(-pka) Ca= initial concentration of the acid = 0.1M The after that you can calculate all the points until the equivalent point with the Henderson Hasselbach equation for a buffer. At the 50% titration (12.5mL) the pH will be equal to the pKa of the acid. then the pH at the equivalence point is going to be a basic salt (potasssium propanoate) an you have to calculate the [OH-] and with the Kb and the concentration of the salt and then calculate the pH

  6. cuanchi
    • one year ago
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    do you need more details?

  7. welshfella
    • one year ago
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    yes you are right I misread the question

  8. cuanchi
    • one year ago
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    |dw:1441917412969:dw|

  9. alphabeta
    • one year ago
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    Oh, thank you SO much @Cuanchi - really appreciating your informative answer! I was just wondering if you mind demonstrating for me Henderson Hasselbach equation? For instance to find out what the pH would be when 5 cm^3 of KOH were added? (And I guess the same equation is used for what you described in your last paragraph?)

  10. cuanchi
    • one year ago
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    Here is a very good complete tutorial. See the example 2 for answer your question. http://chemwiki.ucdavis.edu/Analytical_Chemistry/Quantitative_Analysis/Titration/Titration_Of_A_Weak_Acid_With_A_Strong_Base to solve it you have to do and ICE table . I=initial C= change E= Equilibrium to figure out how many moles of each reactants and products you have at each point of your titration. If you know the number of moles and the volume (that will be changing every time that you add more base) you can calculate the concentrations of the different component of the reaction at any given time. in the Henderson Hasselbach equation used to calculate the [H+] in a buffer system \[[H ^{+}]= pKa+\log \frac{ [base] }{[acid] }\] you dont need to calculate the concentration of the base/acid, because both are in the same volume you can use the moles rate in turn ( the result is the same).

  11. cuanchi
    • one year ago
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    Find the pH after the addition of 5 mL of 0.1 M KOH. SOLUTION 1.- The number of millimoles of propanoic acid to be neutralized is 25mL x 0.1M propanoic acid = 2.5 mmol The number of millimoles of OH- that will be added within 5 mL is 5mL x 0.1 Mol OH− =0.5 mmol 2.- To calculate the pH with this addition of base we must use an ICE Table propanoic acid + OH- -> H2O + propanoate- I-Inital 2.5 mmol 0 mmol - 0 mmol Add 0 mmol 0.5 mmol - Changes -0.5 mmol -0.5 mmol - 0.5 mmol E-Equilibruim 2.0 mmol 0 mmol - 0.5 mmol 3.-This calculation of concentrations can be skipped go to (4) However, this only gives us the millimoles. To get the concentration we must divide by the total volume. The total volume is the 25 mL original solution of propanoic acid plus the 5 mL of KOH that was added. Therefore, the total volume is 25mL+5mL=30mL Concentration of propanoic acid: 2.0 mmol propanoic acid x 30mL=0.12 M Concentration of propanoate - : 0.5 mmol propanoate − x 35mL=0.0175 M Since an acid and its conjugant base are in equilibrium we can attempt to use the Henderson-hasselbalch equation. However, for this to work the reaction must follow certain rules. The ratio of the conjugant base and weak acid must be between 0.10 and 10. Also, both the ratio of the conjugant base and ka value and the ratio of the acid and ka value must exceed 100. In this problem the Henderson-hasselbalch equation can be applied because the ratio of propanoate- to propanoic acid is 0.0175/0.12=0.146. This is between 0.10 and 10. The ratio of propanoic acid to ka is 0.12M/1.32x10−5=9090.9 and the ratio of propanoate- to ka is 0.0175M/1.32x10−5=1325.75. These both exceed one hundred. 4.- Therefore, we continue by using the Henderson-hasselbalch equation. pH=pka+log[A−]/[HA] pH=−log(1.32x10−5)+log(0.5/2.0) pH=4.27

  12. alphabeta
    • one year ago
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    So helpful! Thank you @Cuanchi once again! Should have been able to give more medals than just one for this answer!

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