A certain procedure in lab calls for 400 mL of 0.250 M Tris buffer at pH...

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A certain procedure in lab calls for 400 mL of 0.250 M Tris buffer at pH...

A certain procedure in lab calls for 400 mL of 0.250 M Tris buffer at pH 7.00. You have an available stock bottle of 1.00 M Tris at pH 8.30 and look up the pKa of Tris which is 8.00. You also have stocks of 12.00 M HCl, 12.0 M NaOH, and ddH2O. Which chemicals do you need to add, and what volume?

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Answer 1

Answer #1

[Tris] = x M

[TrisH⁺] =( 0.250 - x)M

pOH = pKb + log((0.250-x)/x)

14 - pH = 14 - 7.00 = 7.00 = 14 - 8.00 + log((0.250-x)/x)

(0.250- x)/x = 10

x = 0.250/11 = 0.02273 M

Millimole of Tris = 0.02273M× 400ml = 100/11

Millimole of TrisH⁺ = (0.250 - 0.02273)×400 = 1000/11

Millimole of HCl added = 1000/11 mmol

Volume of HCl solution taken = (1000/11)÷12.00 = 7.57575ml

Millimole of Tris taken initially = 1000/11 + 100/11 = 100mmol

Volume of Tris solution taken initially = 100mmol /1.00M = 100ml

Amount of water = 400 - (100 + 7.5757 ) = 292.42ml

Answer:

Chemical we needs to add = 7.58ml of 12.00M HCl solution + 100.00ml of 1.00M Tris stock bottle + 292.42ml of distilled water.

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Answer 2

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