Question

2 Introduiction The Henderson-Hasselbalch equation, or A-1 HAl relates the pH of a buffer with the pKo of the acid and the concentration of the conjugate base A- and the monoprotic acid HA. In cq. (1), pH-_ log[H+], pK, =-log Ka, and [J]i and [JIe are the initial and equilibrium molar concentration of the Jth species, respectively. The buffering capacity of the buffer is given by [1] Ka H+ where K,e-1.0 × 10-14 e following In this experiment, you will be assigned a pI by the instructor. Then, answer th You will then use eq. (1) and the information given below to decide how to make the buffer with the pH assigned to you. 1. The buffer will be prepared in a 100 mL volumetric flask. The most concentrated compo- nent in the solution should be held at 0.10 M. Once the chemicals have been approved. calculate the following (a) amounts of each chemical necessary, (b) the equilibrium molar concentrations, (c) the pH at equilibrium, and (d) the buffer capacity 3. Determine the amount of 0.1 M HCl and 0.1 M NaOH you would need to add to a 25 mL sample of your buffer to test the buffer capacity

pH of 9.2

using ammonium nitrate: pKa= 9.24

Answer 1

The concentration ratio of the components of the buffer is calculated by the equation of henderson hasselbach cleared:

[NH3] / [NH4 +] = 10 ^ (pH - pKa) = 10 ^ (9.2 - 9.24) = 0.912

We calculate the concentrations of each component, where the most concentrated is NH4 +:

[NH4 +] = 0.10 M

[NH3] = 0.912 * 0.1 = 0.09 M

a) The moles of each substance are calculated:

n NH4 + = M * V = 0.1 M * 0.1 L = 0.01 mol

n NH3 = 0.09 * 0.1 = 0.009 mol

b) Molar concentrations in equilibrium are:

[NH4 +] = 0.10 M

[NH3] = 0.09 M

c) The pH at equilibrium is 9.2

d) We calculate Ka and the concentration of H +:

Ka = 10 ^ -pKa = 10 ^ -9.24 = 5.75x10 ^ -10

[H +] = 10 ^ -pH = 10 ^ -9.2 = 6.31x10 ^ -10 M

The capacity of the buffer is calculated:

Δn / ΔpH = 2.303 * [(10 ^ -14 / 0.1) + 6.31x10 ^ -10 + (5.75x10 ^ -10 * 6.31x10 ^ -10 / (5.75x10 ^ -10 + 6.31x10 ^ -10) ^ 2 * (0.1 + 0.09)] = 3.65x10 ^ -5

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