Calculate the pH in the tritation of 50.00 mL of 0.060 M acetic acid (CH3COOH) w

Question

Calculate the pH in the tritation of 50.00 mL of 0.060 M acetic acid (CH3COOH) with a 0.120 M sodium hydroxide, NaOH solution after the addition of the following volumes of base: Ka for acetic acid = 1.8 x 10^-5 CHM 1046 Name Homework DUE November 13, 2017 AT 12:30 PM On a separate paper show ALL your calculations and the respective dissociation equations for each step. Assignments without calculations or assignments that are submitted after 12:30 PM on November 13, 2017 are not going to be graded. 1) Calculate the pH in the titration of 50.00 mL of 0.060 M acetic acid (CH,COOH) with a 0.120 M sodium hydroxide, NaOH solution after the addition of the following volumes of base: Ka for acetic acid 1.8 x10 5 A) 0 mL 7 B) 12.5 mL C) 25.0 mL D) 30.0 ml

Explanation / Answer

Moles of acetic acid = Volume* molarity

Moles of acetic acid = 0.060*50/1000 =0.003

The reaction between acetic acid ( represented as HA) and NaOH is

HA+ NaOH---àNaA+ H2O

1 mole of HA requires 1 mole of NaOH. So molar ratio of HA: NaOH= 1:1 ( theoretical)

a) When no NaOH is added,

HA undergoes ionization as HA+ H2O---àA- + H3O+

Ka= [A-] [H3O+]/ [HA]

Writing ICE Table

b)When 12.5 ml of 0.12M NaOH is added, moles of NaOH added = 0.12*12.5/1000 =0.0015

Moles of acetic acid supplied =0.003, acetic acid to NaOH molar ratio (actual) =0.003:0.0015= 2:1

So excess isHA and all the NaOH is consumed.

Hence moles of NaA formed =0.0015 , moles of acetic acid remaining = 0.003-0.0015

Volume of solution after mixing = 12.5+50= 62.5ml, 1000ml= 1L, 62.5ml= 62.5/1000L= 0.0625L

Concentrations : NaA= 0.0015/0.0625 M and HA= 0.0015/0.0625

pH= 4.74+ log (0.0015/0.0015)= 4.74

c)) Moles of NaOH in 25ml= 0.12*25/1000 =0.003, moles of acetic acid = 0.003

So all the moles of acetic acid and NaOH gets consumed andmoles ofsodium acetate formed =0.003. This is the equivalence point.

Volume of solution after mixing = 25+50= 75ml= 75/1000L=0.075 L

Concentration ofNaA after mixing = 0.003/0.075 M=0.04M

Sodium acetate (NaA isionized as )

A-+ H2O---->HA+ OH-

Kb= 10-14/Ka= 5.55*10-10 =[HA] [OH-]/[A-]

Writing ICE Table

d) When 30 ml of 0.12M NaOH is added, mole of NaOH=0.12*30/1000 =0.0036, mole of acetic acid =0.003

This suggests that moles of NaOH is excess and moles of NaOH remaining after reaction with HA is = 0.0036-0.003= 0.006

Volume of solution after mixing = 30+50= 80ml =80/1000L= 0.08L

Concentration of NaOH= moles/ Volume in L= 0.006/0.08= 0.0075

NaOH ionizes as NaOH ---à Na++ OH-

[OH-]= 0.0075, pOH= -log (0.0075)= 2.12, pH= 14-2.12= 11.88

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