Question

For 460.0 mL of a buffer solution that is 0.125 M in HC2H3O2 and 0.110 M in NaC2H3O2, calculate the initial pH and the final pH after adding 0.010 mol of HCl.

For 460.0 mL of a buffer solution that is 0.165 M in CH3CH2NH2 and 0.155 M in CH3CH2NH3Cl, calculate the initial pHand the final pH after adding 0.010 mol of HCl.

Answer 1

v= 460ml

pH = pKa + log (salt/acid)

pH=4.47 + log (0.11/0.125) = 4.47- 0.055=4.415

HC2H3O2 --> H+ + C2H3O2-
Ka = 1.8x10^-5 = x^2 / 0.125 - x
x^2 + 1.8x10^-5x - 2.25x10^-6 = 0
H+ =1.49 x10^-3M

C2H3O2- --> HC2H3O2 + OH-
Kb = 5.56x10^-10 = x^2 / 0.11M
x^2 = 6.117x10^-11
x = [OH-] = 7.821x10^-6M

mol of OH- = 7.821x10^-6 x 0.46 =3.59766x10^-6mole

0.01moles H+ will neutralized all of the OH- produced by hydrolysis

0.01- 0.00000358 = 9.99x10^-3 mol H+ left

1.49x10^-3 x.46 mole H+ from acid = .68x10^-3 mol

total mole of H= 9.99x10^-3 + .68x10^-3 = 10.675x10^-3 mol

pH = - log [H+] = - log 10.675x10^-3/.46= -log 0.0232=1.6345

b-

Kb of ethylammine = 4.3 x 10^-4
pKb = 3.37

pOH = pKb +log salt/base = 3.37 + log (0.155/0.165) = 3.37-0.027= 3.343

pH= 14- 3.343=10.657

moles of CH3CH2NH3+=0.155x0.46=0.0713

moles of CH3CH2NH2 = 0.165x.46 = 0.0759

CH3CH2NH2+H+-----> CH3CH2NH3+

mole of CH3CH2NH3+= 0.0713+ 0.01=0.0813 conc. = 0.0813/.46=0.1767

mole of CH3CH2NH2= 0.0759-0.01= 0.0659     conc.= 0.0659/0.46 = 0.14326

pOH= 3.37+log(0.1767/0.14326)= 3.37+0.90= 4.27

pH=14-4.27=9.73