Calculate the ph for each of the following cases in the titration of 25.0 ml of 0.150 m acetic acid (ka = 1.75x10-5) with 0.150 m naoh(aq). (1 point each) (a) before addition of any naoh (b) after addition of 5.0 ml of naoh (c) after addition of 15.0 ml of naoh (d) after addition of 25.0 ml of naoh (e) after addition of 40.0 ml of naoh (f) after addition of 60 ml of naoh

a) pH = 2.793

b) pH = 4.280

c) pH = 4.933

d) pH = 8.816

e) pH = 8.861

f) pH = 8.891

Explanation:

a) VNaOH = 0 mL

∴ CH3COOH ↔ CHECOO- + H3O+

⇒ Ka = 1.75 E-5 = [ H3O+ ] * [ CH3COO-] / [ CH3COOH ]

mass balance:

⇒ C CH3COOH = [ CH3COO- ] + [ CH3COOH ] = 0.150 M

charge balance:

⇒ [ H3O+ ] = [ CH3COO- ]

⇒ Ka = 1.75 E-5 = [ H3O+ ]² / ( 0.150 M - [ H3O+ ] )

⇒ [ H3O+ ]² + 1.75 E-5 [ H3O+ ] - 2.625 E-6 = 0

⇒ [ H3O+ ] = 1.61146 E-3 M

⇒ pH = - Log [ H3O+ ] = 2.793

b) after  5.0 mL NaOH:

∴ CH3COOH + NaOH ↔ CH3COONa + H2O

⇒ C NaOH = (5 E-3 L * 0.150 mol/L) / (0.025+0.01 ) = 0.02143 M

⇒ C CH3COOH = ((0.025*0.150) - (0.01*0.150)) / (0.025 + 0.01) = 0.0643 M

mass balance:

⇒ 0.02143 + 0.0643 = [ CH3COOH ] + [ CH3COO- ] = 0.086 M

charge balance:

⇒ [ H3O+ ] + [Na ] = [ CH3COO- ]

⇒ [ H3O+ ] + 0.02143 = [ CH3COO- ]

⇒ Ka = [ H3O+ ] * ( [ H3O+ ] + 0.150 ) / (0.086 - 0.02143 - [ H3O+ ]) = 1.75 E-5

⇒ [ H3O+ ]² + 0.02143 [ H3O+ ] = 1.13 E-6 - 1.75 E-5 [ H3O+ ]

⇒ [ H3O+ ]² + 0.02144 [ H3O+ ] - 1.13 E-6 = 0

⇒ [ H3O+ ] = 5.26 E-5 M

⇒ pH = 4.28

c) after 15 mL NaOH:

⇒ C CH3COOH = 0.0375 M

⇒ C NaOH = 0.05625 M

mass balance:

⇒ 0.09375 M = [ CH3COO- ] +[ CH3COOH ]

charge balance:

⇒ [ H3O+ ] + 0.05625 = [ CH3COO- ]

⇒ Ka = 1.75 E-5 = [ H3O+ ] * ([ H3O+ ] + 0.05625) / (0.09375 - 0.05625 - [H3O+])

⇒ [H3O+]² + 0.05625[H3O+] = 6.5625 E-7 - 1.75 E-5 [H3O+]

⇒ [ H3O+]² + 0.05626[H3O+] - 6.5625 E-7 = 0

⇒ [ H3O+ ] = 1.1662 E-5 M

⇒ pH = 4.933

d) after 25 mL NaOH:

⇒ C NaOH = 0.075 M

⇒ C CH3COOH = 0 Mequiv. point

⇒Kh = Kw/Ka = 1 E-14 / 1.75 E-8 = 5.7143 E-10 = [ OH-]² / ( 0.075 - [OH-])

⇒ [OH-]² + 5.7143 E-10[OH-] - 4.286 E-11 = 0

⇒ [ OH- ] = 6.5463 E-6 M

⇒ pOH = 5.184

⇒ pH = 8.816

e) after 40 mL NaOH:

⇒ C NaOH = 0.0923 M

⇒ [OH-]² + 5.7143 E-10 [OH-] - 5.275 E-11 = 0

⇒ [OH-] = 7.2624 E-6 M

⇒ pOH = 5.139

⇒ pH = 8.861

f) after 60 mL NaOH:

⇒ C NaOH = 0.106 M

⇒ [OH-]² + 5.7143 E-10 [OH-] - 6.05 E-11 = 0

⇒ [OH-] = 7.7782 E-6 M

⇒ pOH = 5.11

⇒ pH = 8.891

i believe the answer is a.

56 moles of   lieter in chemistry