1 10 points Consider the titration of 1000 mL of 0100 M meth

1. (10 points) Consider the titration of 100.0 mL of 0.100 M methylamine (CH,NH2) with o ton HNO3. Calculate the pH at the following volumes of acid added. For CHaNHs, pKa 10.632 (a) Find the equivalence point volume. (b) 0 mL (c) 9.0 mL (dy lo o mi F) 36.oml

Solution

a)At equivalence point

mol of CH3NH2=0.1L*0.1mol/L=0.01mol

mol of HNO3=mol of CH3NH2=0.01mol

CH3NH2 +HNO3 --->CH3NH3⁺ +NO3- (neutralization)

mol of CH3NH3⁺ =0.01mol

volume of HNO3 added=mol of HNO3/molarity =0.01mol/0.5mol/L=0.02 L=20ml

[CH3NH3+]=0.01mol/120ml=0.01mol/0.120L=0.0833mol/L

So at equilibrium only the salt CH3NH3⁺ remains in the solution,that hydrolyses according to th equation:

CH3NH3⁺ +H2O <---> CH3NH2 +H3O+

ka=[CH3NH2] [H3O+ ]/[CH3NH2]

pka=10.632=-log ka

ka=10^-pka=10^-10.632=2.333*10^-11

2.333*10^-11=[CH3NH2] [H3O+ ]/[CH3NH3+]

ICE table

2.333*10^-11=x^2/(0.0833-x) [x<<0.0833 as very low dissociation takes place]

2.333*10^-11=x^2/(0.0833)

or,x=1.394*10^-6M=[H3O+]

pH=-log [H3O+]=-log (1.394*10^-6M)=5.8

pH=5.8

2) 0.0ml HNO3 added

only CH3NH2 in solution that hydrolyses to give

CH3NH2+H2O <--->CH3NH3+ +OH-

kb=kw/ka=(10^-14)/(2.333*10^-11)=0.000428

kb=[CH3NH3+] [OH-]/[CH3NH2]

preparing ICE table as in part a)

kb=0.000428=x^2/(0.1-x)

solving for x,

x=0.00654 M=[OH-]

[H3O+][OH-]=10^-14(=kw(ionic product of water)

[H3O+]=10^-14/(0.00654)=1.529*10^-12M

pH=-log (1.529*10^-12M)=11.8

3) 9ml=0.009L of HNO3 added

mol of HNO3 added=0.009L*0.5mol/L=0.0045mol

CH3NH2 +HNO3 --->CH3NH3⁺ +NO3- (neutralization)

So mol of CH3NH3⁺ formed=0.0045mol

[CH3NH3⁺ ]=0.0045mol/(109ml)=0.0045mol/(0.109L)=0.0413M

mol of CH3NH2 remaining=0.01mol-0.0045mol=0.0055mol

[CH3NH2]=0.0055mol/0.109L=0.0504M

pH=pka+log [base]/[acid](henderson -hasselbach equation)

pH=10.632+log [CH3NH^2]/[CH3NH3^+] =10.632+log (0.0504/0.0413)=10.7

d)10ml=0.010Lof HNO3 added

mol of HNO3 added=0.010L*0.5mol/L=0.005mol

CH3NH2 +HNO3 --->CH3NH3⁺ +NO3- (neutralization)

So mol of CH3NH3⁺ formed=0.005mol

[CH3NH3⁺ ]=0.005mol/(109ml)=0.005mol/(0.109L)=0.0459M

mol of CH3NH2 remaining=0.01mol-0.005mol=0.005mol

[CH3NH2]=0.005mol/0.109L=0.0459M

pH=pka+log [base]/[acid](henderson -hasselbach equation)

pH=10.632+log [CH3NH^2]/[CH3NH3^+] =10.632+log (0.0459/0.0459)=10.632

pH=pka(half -neutralization)

	[CH3NH3+]	[CH3NH2]	[H3O+ ]
initial	0.0833M	0	0
change	-x	+x	+x
equilibrium	0.0833-x	x	x