Question

(3 pts) 4. You are asked to prepare a 0.100 M)glycine buffer pH 9.0. You are given a bottle of glycine (FW 75.1 g/mol) and a stock solution of 1.00 M NaOH. How many grams of glycine and how many milliliters of NaOH stock solution would be required to produce 2 liters of this buffer solution? pk,95 alyeine tr oon 2 38 CHA) 7 3A1 (HA o.l0o glycine 3.98

Answer 1

By means of the hasselbach henderson equation, the molar ratio of the salt formed by the addition of NaOH and acid is calculated:

n Salt / n Acid = 10 ^ (pH-pKa) = 10 ^ (9 - 9.6) = 0.25

It clears:

1) n Salt - 0.25 * n Acid = 0

The total glycine moles are:

2) n Gly = n Salt + n Acid = M * V = 0.1 M * 2 L = 0.2 mol

With 1 and 2 a system of equations is made and we have:

n Salt = 0.04 mol

n Acid = 0.16 mol

The grams of glycine are calculated:

g Gly = n Gly * MM = 0.2 mol * 75.1 g / mol = 15.02 g

The mL of NaOH that forms the salt are calculated:

mL NaOH = n * 1000 / M = 0.04 mol * 1000/1 M = 40 mL

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