Question

A 25.0 mL sample of 0.203 M ethylamine, C2H^NH2, is titrated with 0.225 M nitric acid. Before the addition of any nitric acid, the pH is (2) After adding 8.62 mL of nitric acid, the pH is (3) At the titration midpoint, the pH is (4) At the equivalence point, the pH is (5) After adding 35.4 mL of nitric acid, the pH is Use the Tables link on the toolbar for any equilibrium constants that are required. Not submitted

Answer 1

1-Before the addition of any HNO3 0.203 M 0 nitial 0.203 M-x x equilibrium [C2H5 NH江HO-1-63 x 10-4 0.203-= 6.3 x 10-4 0.203-x-6.3 x 10-4 kb is very small so that, x can be depreciated. 0.203 6.3x 104 x0.203 x 6.3 x 10-4

[HO]-x 0.0113 M pOH =-log[HO-] = _log0.0113 M = 1.95 14 = pH + pOH 14 = pH + 1.95 14-1.95 pH 12.05 = pH 2. Calculate the initial moles of HNO3 and CH5NH2 0.203 mol C2H5NH2 1000 ml initial moles C2H5NH2 25.0 mL x 0.00508 mol C2HsNH2 0.225 mol HNO3 1000 ml initial moles HNO2 8.62 mL x = 0.00194 mol HNO3 Calculate the final [C2HsNH2 ] and [C2HsNH3] mol 1 mol 1 mol 1 mol 1 mol C2H5NH2 1 mol HNO3 moles C-H5NH2 that react = 0.00194 mol HNO3 x = 0.00194 mol C2H5NH2

moles C2HsNH produced 0.00194 mol HNO3 x 0.00194 mol C2HsNH 1 mol C2HsNHj mol HNO3 final moles C2H5NH20.00508 mo 0.00194 mol 0.00314 mol C2H5NH2 0.00194 mol C2HsNH31000 mL x- 0.0577 M 25.0 mL 8.62 mL 1 L 0.00314 mol C2HsNH2 1000 mL x 0.0934 M 25.0 mL8.62 mL 1L Calculate the pH [base] [acid] 0.0934 M H = 10.75 + log 0.0577 M pH- 10.75 + 0.209 pH 10.96

In the midpoint, So that, pH 10.75 log1 pH-10.75 In the equivalence point all the weak base was neutralized by the strong acid and the pH depends of the dissociation of the conjugate acid in the following way moles C2HsNH initial moles C2HsNH2 0.00508 mol C2HsNH2 Calculate the acid volume added in the equivalence point 1 mol 1 mol 1 mol mo 1 mol HNO3 1 mol C2HsNH necessary moles HNO3 0.00508 mol C2HsNH^ x 0.00508 mol HNO

1000 mL volume HNO3 = 0.00508 mol HNO3 x = 22.6mL 0.225 mol HNO3 So that, 0.00508 mol C2HNH2 1000 mL x = 0.107 M 25.0 mL 22.6 mL 1 L Calculate the pH 0.107 M 0 ntial 0.107 M-x x equilibrium [c2H5NH+]-= 1.78 x 10-11 = 1.78 x 10-11 0.107-x = 1.78 x 10-11 0.107 x ka is very small so that, x can be depreciated.

一7 = 1.78 x 10-11 x = 0.107 x 1.78 x 10-11 [H11.38x 10-5 M pH--loglr]--l0g1.38 10* M -5.86