A weak acid with an initial pH of 3.2 was titrated with a strong base. 15 mL of 0.1 M NaOH was added to the acid to reach the equivalence point at a pH of 8.6. What would you expect the approximate pH of the analyte to be after the first 5 mL of 0.1 M NaOH was added? A. 5.1 B. 7.2 C. 3.4 D. 8.6
Do you have a guess of which one it is?
I got 3.27
So maybe it's C?
I'm thinking so I don't think I messed up on my math would you like to know what I did?
Ok so we have 15ml of 0.1M NaOH. We can find the moles of NaOH by multiplying the volume in liters (0.015L) by the molarity of 0.1
So that means I have 0.0015moles of NaOH
Then we can set up the ice table HA+NaOH ---> NaA+ H2O with the initial moles for NaOH as the 0.0015moles
I am really bad at explaining this I followed what they did in this link http://chemwiki.ucdavis.edu/Analytical_Chemistry/Quantitative_Analysis/Titration/Titration_Of_A_Weak_Acid_With_A_Strong_Base
Okay, hold on let me read it.
Let me try this again and I'll let you know
Now I'm getting an answer no where near anything my answer was just over one which does not make sense one bit!
I don't think you need to do any math for the question since all the possible options are so far apart. Picture a titration curve |dw:1437179500727:dw| Initially, and for most of the curve, the slope is small, so we don't expect use changes in pH. If you used 1/3 of the moles that got you to 8.6 (which is a difference of about 5 from 3.2). so, 1/3 of 5 is 1.7, so 3.2 + 1.7\(\approx\)5 so I think 5.1 is reasonable
sorry my graph is terrible and unrepresentative of whats going on lol
that makes sense just to clarify though the 1/3 you get from the fact that it's 5mL out of the 15mL?
yeah
Sounds good it makes a lot more sense than the number I got and not having to do all that math is even better lol
the math is often needed but it doesnt always have to be the first way to approach a problem.
It was 3.4
Good to know!
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