Question

In: Chemistry

For the titration of 50.0 mL of 0.150 M acetic acid with 0.100 M sodium hydroxide,...

For the titration of 50.0 mL of 0.150 M acetic acid with 0.100 M sodium hydroxide, determine the pH when:

(a) 50.0 mL of base has been added.

(b) 75.0 mL of base has been added.

(c) 100.0 mL of base has been added.

Solutions

Expert Solution

moles of acetic acid in 0.15 M of 50ml =0.15*50/1000 =0.0075

moles of sodium hydroxide 50 ml of 0.1M=0.1*50/1000=0.005

the reaction between acetic acid and sodium hydroxide is CH3COOH+NaOH -------->CH3COONa+ H2O

molar ratio of CH3COOH: NaOH= 1:1

actual ratio of CH3COOH: NaOH=0.0075:0.005

excess is CH3COOH and limiting is Sodium hydroxide. All the moles of sodium hydroxide is consumed. moles of sodium acetate formed =0.005, moles of acetic acid remaining =0.0075-0.005 =0.0025

volume after mixing= 50+50=100ml =(100/1000)L =0.1L

concentrations (M): acetic acid = 0.0025/0.1, sodium acetate = 0.005/0.1

since pH= pKa+ log [A-]/[HA]

[HA] is acetic acid and [A-] is sodium acerate

Pka of acetic acid = 4.76, Ka= 10(-4.76)= 1.74*10-5

pH= 4.76+ log (0.005/0.0025)=5.06

2. moles in 75 ml of 0.1M= 0.1*75/1000 = 0.0075 . moles of acetic acid also=0.0075

this is equivalence point, moles of sodium acetate formed =0.0075

volume of solution after mixing = 50+75=125ml= 125/1000L=0.125L

concentration of sodium acetate after mixing = 0.0075/0.125 =0.06 M

CH3COO- + H2O ---------->CH3COOH+ OH-

Kb= [CH3COOH] [H3O+]/[CH3COO-]= 10-14/Ka= 10-14/1.74*10-5= 5.75*10-10

preparing the ICE Table

component                       CH3COO-                                 [OH-]              [CH3COOH]

initial                                   0.06                                           0                           0

change                                 -x                                              x                          x

Equilibrium                        0.06-x                                         x                           x

x2/ (0.06-x)= 5.75*10-10 , when solved using excel, x= 5.87*10-6

pOH= -log [OH-]= -log (5.87*10-6)= 5.23, pOH= 14-pOH= 14-5.23= 8.77

3 moles of NaOH in 100 ml of 0.1M =0.1*100/1000 =0.01moles, moles of acetic acid =0.0075

excess is NaOH and all the acetic acid gets consumed. Moles of NaOH remaining =0.01-0.0075= 0.0025

volume of solution after mixing = 100+50= 150 ml= (150/1000)L, concentration of NaOH remainig =0.0025*1000/150 =0.017 Moles/L

since NaOH is strong base, it ionizes completely. NaOH ------->Na++ OH-

pOH= -log (0.017)= 1.78

pH= 14-1.78=12.22


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