3. Consider the titration of 100.0 ml of 0.200 M CH:COOH (Ka 1.8x10) by 0.100 M

Question

3. Consider the titration of 100.0 ml of 0.200 M CH:COOH (Ka 1.8x10) by 0.100 M KOH. a. Write the net ionic equation for this titration. b. Identify a conjugate acid-base pair in the above reaction, labeling the acid and the base. c. Calculate the pH after 100.0 ml of KOH has been added to the given solution. Calculate the pH of the resulting solution after 200.0 ml of KOH have been added. What is this point in the titration called? d. 4. Calculate the pH of a solution that is 0.600 M NH3 and 1.00 M NHABr. (K,-1.8 x 10-5). This type ofsolution is called a 5. Fill in the blanks in the following table: pH pOH HT OH] Acid, base, or neutral? Solution A 6.88 Solution B Solution C 3.4 x 10-13 1.0 x 107

Explanation / Answer

3a) Balanced rxn for neutralization:

CH3COOH(aq) +KOH(aq) --->CH3COOK(aq) +H2O(l)

ionic eqn: CH3COO-(aq) +H+(aq) K+(aq) + OH-(aq) ---->CH3COO-(aq) +K+(aq) +H2O(l)

removing all spectator ions,

net ionic eqn: H+(aq) + OH-(aq) ---->H2O(l)

b) conjugate acid : CH3COOH ,conjugate base:CH3COO- (acetate)

conjugate acid :H+ conjugate base:OH-

c) mol of KOH added=0.1mol/L*0.1L=0.01mol

mol of Ch3COOH=mol of KOH=0.01mol

total mol of CH3COOH=0.2mol/L*0.1L=0.02mol

remaining mol of CH3COOH after rxn with KOH=0.02mol-0.01mol=0.01mol

Using henderson -hasselbach eqn,

pH=pka+log[CH3COO-]/[CH3COOh]

pka=-logka=-log (1.8*10^-5)=4.7

pH=4.7+log (0.01/0.01)=4.7

pH=pka=4.7(half -equivalence point as [base]=[acid]

d) mol of KOH added=0.1mol/L*0.2L=0.02mol

mol of Ch3COOH=mol of KOH=0.02mol

total mol of CH3COOH=0.2mol/L*0.1L=0.02mol

remaining mol of CH3COOH after rxn with KOH=0.02mol-0.02mol=0 mol

This point iscalled the equivalence point ,only CH3COO- isin solution,

mol of CH3COO-=0.02mol

[CH3COO-]=0.02mol/300ml=0.02mol/0.3L=0.0667 M

Hydrolysis of CH3COO- takes lace,

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

kb=kw/ka=10^-14/(1.8*10^-5)=5.555*10^-10

kb=[CH3COOH][ OH-]/[CH3COO-]

ICE table

kb=x^2/(0.0667 M-x) [x<<<0.0667M,so can be ignored]

5.555*10^-10=x^2/0.0667

x=[OH-]=6.087*10^-6M

pOH=-log [OH-]=-log (6.087*10^-6)=5.21

pH=14-pOH=14-5.21=8.8

pH=8.8

[CH3COO-] [CH3COOH] [OH-] initial 0.0667 M 0 0 change -x +x +x equilibrium 0.0667 M-x x x

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