I need help with parts C, D, and E. I provided info from parts A and B in case t

Question

I need help with parts C, D, and E. I provided info from parts A and B in case they are needed to calculate the rest of the problems. Thank you!

Part D Constants Periodic Table Consider the titration of 37.6 mL of 0.260 M HF with 0.200 M NaOH. Calculate the pH at each of the following points. Calculate the pH at the equivalence point. Express your answer using two decimal places. pH 7,0 Submit Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining Part E Calculate the pH after the addition of 75.2 mL of base. Express your answer using two decimal places. pH- Submit Request Answer

Explanation / Answer

Given :M(HF)=0.26M

Volume(HF)=V(HF)=37.6ml=37.6ml*(1L/1000ml)=0.0376L

mol of HF=molarity*volume=0.26 mol/L*0.0376L=0.00978 mol

part D)

mol of NaOH to be added at equivalence point=mol of HF=0.00978mol

NaOH +HF-->NaF+H2O

mol of NaF formed=0.00978 mol

volume of NaOH added=mol NaOH/M(NaOH)=0.00978 mol/0.2mol/L=0.0489 L=48.9ml

total volume of solution=48.9ml+37.6ml=86.5 ml=0.0865L

[NaF]=mol/total volume=0.00978 mol/0.0865L=0.113 M

Now there is only NaF in the solution that dissociates completely to give F-, and F- hydrolyses in solution:

NaF--->Na+ +F-

F- +H2O --->HF +OH-

Kb=base dissociation constant=[HF][OH-]/[F-]

Acid dissociation constant, Ka of HF=7.2*10^-4

Kb=Kw/Ka=(1*10^-14)/(7.2*10^-4)=1.389*10^-11 [Kw=ionic product of water]

ICE table

Kb=1.389*10^-11 =x^2/(0.113-x) [x<<<0.113M,for kb being very small)

1.389*10^-11 =x^2/(0.113)

solving for x,

x=[OH-]=1.253*10^-6M

[H3O+][H+]=kw

[H+]=(10^-14)/(1.253*10^-6)=7.981*10^-9M

pH=-log[H+]=-log (7.981*10^-9M)=8.1

pH=8.1

part C)

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

or,pH=pka +log [F-]/[HF]

At half equivalence , Amount acid neutralized=1/2 of initial amount=1/2[HF]o

So, mol of F- formed=1/2[HF]o

mol of HF remaining=1/2[HF]o

[F-]=[HF] at half-equivalence

pH=pka+log (1)

or pH=pka=-log Ka=-log(7.2*10^-4)=3.1

pH=3.1=pka

part E)

After 75.2ml of NaOH is added,mol of NaOH added=0.2mol/L*0.0752L=0.0150 mol

mol of HF=molarity*volume=0.26 mol/L*0.0376L=0.00978 mol

Excess mol of OH-=0.0150 mol-0.00978 mol=0.00522 mol

total volume=75.2ml+37.6ml=112.8ml=0.1128L

[OH-]=0.00522 mol/0.1128L=0.0461 mol/L

[H3O+]=1*10^-14/[OH-=(1*10^-14)/0.0461=2.169*10^-13M

pH=-log [H3O+]=-log (2.169*10^-13M)=12.7

pH=12.7

[F-] [HF] [OH-] initial 0.113M 0 0 change -x +x +x equilibrium 0.113M-x x x

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