Consider the titration of 50.0ml of 0.0200M diprotic acid with pk1=4.00 and pk2=8.00 by 0.100M NaOH....

Consider the titration of 50.0ml of 0.0200M diprotic acid with pk1=4.00 and pk2=8.00 by 0.100M NaOH. Calculate the pH at the following points: 0ml, 2ml, 5ml, 7ml, 10ml, 15ml, 20ml, 25ml

Expert Solution

queen_honey_blossom answered 2 years ago

Titration of 50.0mL of 0.100M HX (Ka=1.5x10^-5) with 0.100M NaOH. Calculate the pH of: - buffer...

Titration of 50.0mL of 0.100M HX (Ka=1.5x10^-5) with 0.100M NaOH. Calculate the pH of: - buffer formed at the addition of 12.5mL NaOH - buffer formed at the addition of 25.0mL NaOH - buffer formed at the addition of 37.5mL NaOH - solution obtained at the endpoint Please do the endpoint especially!

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sketch the general appearance of the curve for the titration of weak diprotic acid with NaOH....

sketch the general appearance of the curve for the titration of weak diprotic acid with NaOH. what chemistry governs the pH in each distinct region of the curve

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50.0ml of 0.125M HF is titrated with 0.100M NaOH. Calculate the pH after the addition of a) 0.00ml of NaOH b) 15.00ml of NaOH c) 31.25ml of NaOH d) 60.00ml of NaOH e) 62.50ml of NaOH f) 70.00ml of NaOH

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titration of 25.00 ml of 0.100M CH3COOH with 0.100 M NaOH (weak acid, strong base) a) calculate the initial pH ( kb = 1.8 x10 ^-5) b) why is pH > 7 at the equivalence point? c) calculate the pH at the equivalence point

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For the triprotic acid, H+glutamic acid, pK1=2.23, pK2=4.42 and pK3=9.95. If one has 0.10L of a 1.0E-2 M solution and titrates it with 0.255 L of a 1.0E-2 M KOH solution, what is the pH of the resulting solution? The answer is 10.04. I just have no clue how to get there.

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Calculate the pH in the titration of 25.0 mL of 0.100M HCl with 0.200M NaOH. Finally, calculate the pH after 18.75 mL of base have been added.

Calculate the pH in the titration of 25.0 mL of 0.100M HCl with 0.200M NaOH. Next,...

Calculate the pH in the titration of 25.0 mL of 0.100M HCl with 0.200M NaOH. Next, calculate the pH after 6.25 mL of base have been added.

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A 25.00 mL aliquot of a 0.100 M maleic acid, HOOC-CH=CH-COOH (pK1=1.87) and (pK2 = 6.23) was titrated with 0.1000 M NaOH. Construct a titration curve by calculating pH at a) initial pH b) after adding 5.00 mL of NaOH c) after adding 24.90 mL of NaOH d) at equivalence point e) after adding 25.01 mL of NaOH f) after adding 25.50 ml of NaOH g) after adding 49.90 mL NaOH h) after adding 50.00 mL NaOH i) after adding...

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