If you really wnated to prepare a tris buffer of pH 7.60, you would not do it by

Question

If you really wnated to prepare a tris buffer of pH 7.60, you would not do it by calculating what to mix. Suppose that you wish prepare 100mL of buffer containg 0.100M tris at a pH of 7.60. You have available solid tris hydrochloride and approximately 1.0 M NaOH. If you really wnated to prepare a tris buffer of pH 7.60, you would not do it by calculating what to mix. Suppose that you wish prepare 100mL of buffer containg 0.100M tris at a pH of 7.60. You have available solid tris hydrochloride and approximately 1.0 M NaOH. Suppose that you wish prepare 100mL of buffer containg 0.100M tris at a pH of 7.60. You have available solid tris hydrochloride and approximately 1.0 M NaOH.

Explanation / Answer

Let x ml of 1.0 M NaOH is added.

so x ml of 1.0 M NaOH = x * 1.0 / 1000 = 0.001 x mole.

resulting solution is basic buffer solution and TRIS is weak base.

we have 0.01 mol of tis hydrochoride.

moles of base = 0.001x mole.

moles of salt = (0.01 - 0.001x) moles.

PKa of  tis hydrochoride = 8.075

PKa + PKb = 14

PKb = 14 - 8.075 = 5.925

POH = 14 - PH = 14 - 7.60 = 6.4

we know,

POH = PKb + log [salt] / [base]

6.40 = 5.925 + log (0.01 - 0.001x) / 0.001 x

(0.01 - 0.001x) / 0.001 x = 2.9854

x = 2.51 ml

so 2.51 ml NaOH is added.

100 ml of 0.100 M tris hydrochloride = 0.100 * 0.100 = 0.01 mole = 0.01 * 157.59 = 1.58 gm.

so to prepare 100 ml of 0.1000 M TRIS buffer, we mixed 2.51 ml 1.0 M NaOH , 1.58 gm tris hydrochloride and (100 - 2.51) = 97.49 ml water.

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