How many mL of 0.500M NaOH should be added to 12.114 g of “tris” hydrochloride (

Question

How many mL of 0.500M NaOH should be added to 12.114 g of “tris” hydrochloride (HOCH2)3NH3+Cl- (MW=121.14 g/mol, whose pKa= 8.07) to reach pH=7.92 in a final volume of 100 mL?

Assume that “x” moles of NaOH is added, complete the following table. (Note that you are allowed to use “x” in the expression only in this table)

OH-

(HOCH2)3NH3+

(HOCH2)3NH2

H2O

Initial moles

x mol

Y

0 mol

Final moles

0 mol

(0.10000-x)mol

Or 0.059

X

Or 0.041

Give the value of “Y”:_ 0.10000 mol

Give the value of “X”:_0.041 mol

Give the volume of 0.500M NaOH: 0.083L

*******My questions are, where did the answers 0.041 and 0.083 come from? Please show work.

OH-

(HOCH2)3NH3+

(HOCH2)3NH2

H2O

Initial moles

x mol

Y

0 mol

Final moles

0 mol

(0.10000-x)mol

Or 0.059

X

Or 0.041

Explanation / Answer

let x =moles of NaOH added

moles of tris buffer = 12.114/112.114= 0.1 moles

OH- + (HOCH2)3 NH3+ ------------(HOCH2)3NH2 + H2O

since all the x moles are consumed, moles of tris buffer remaining at equilibrium = 0.1-x

moles of (HOCH2)3NH2 formed = x

pH= pKa + log[conjugate base/ acid]

7.92 =8.07 + log [B/A]

B/A =0.71

x/(0.1-x)= 0.71

x = 0.071-0.71x

1.71x = 0.071, x = 0.041, hence A =0.1-0.041= 0.059 and base =0.059

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