Sulfanilic acid(NH 2 C 6 H 4 SO 3 H) is anorganic compound used in the manufactu

Question

Sulfanilic acid(NH2C6H4SO3H) is anorganic compound used in the manufacture of dyes. It ionizesin water according to the equilibrium equation.

NH2C6H4SO3H(aq)+ H2O(l) (<Reaction Arrow>)NH2C6H4SO3-(aq)       Ka = 5.9 x 10-4

A buffer is prepared by dissolving 0.20 mol of sulfanilicacid and 0.13 mol of sodium sulfanilate
NH2C6H4SO3 in water anddiluting it to 1.00 L.

a) compute the pH of the solution.

b) Suppose that 4.00 mL of 0.010 M HCI is added to thebuffer. Calculate the pH of the solution that results.

Explanation / Answer

            NH2C6H4SO3H(aq) + H2O(l) NH2C6H4SO3-(aq)   +     H3O+ initial               0.2                                                0.13                                0 change              -x                                                   +x                               +x equil               0.2-x                                              0.13 +x                          x Ka = [H3O+][NH2C6H4SO3-]/[NH2C6H4SO3H] = x*(0.13+x)/(0.2-x) =5.9e-4 x2 + (0.13 +5.9e-4)x - 0.2*5.9e-4 = 0 Solve for x. by quadratic formula x = 0.000897424227 pH = -log[H3O+] = -log(x) = 3.04700221 ~ 3.05 b) 4 mL *(1 L/1000 mL) *(0.010 mol H+/1L) = 4e-5 total volume = 1.004 Liters now initial acid concentration = (0.20 mol)/1.004 L = 0.199203187 M initial conjugate base concentration = 0.13/1.004 = 0.129482072M initial H3O+ concentration = 4e-5/1.004 L = 3.98406375e-5 M set up equilibrium :             NH2C6H4SO3H(aq) + H2O(l) NH2C6H4SO3-(aq)   +     H3O+ initial               0.1992                                            0.12948                          3.984 e-5 change              -x                                                   +x                                +x equil               0.1992-x                                       0.12948 +x                     3.984e-5 + x Ka = 5.9e-4 = (3.984e-5 + x)*(0.12948 + x)/(0.1992 - x) x2 + (3.984e-5 + 0.12948)*x + (3.984e-5)*0.12948 =5.9e-4*0.1992 - 5.9e-4*x x2 + (3.984e-5 + 0.12948+5.9e-4)*x +(3.984e-5)*0.12948 -5.9e-4*0.1992 = 0 x2 + 0.130111913*x -0.000112371232 = 0 Solve for x via quadratic formula x = 0.000857992768 M Final [H3O+] = 3.984e-5 + x = 0.000897833406 M pH = -log[H3O+] = 3.04680424 ~ 3.05 Notice that pH did not change much. This is the benefit of using abuffer

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