Question

Consider the titration of 50.0 mL of 0.100 M HN3 (Ka=1.9X10^-5) with 0.200 M CSOH. Calculate the PH after addition of

A) at initial point

B) 12.50 ml CSOH

C) 50.0 ml CSOH

D) 60.0 mL of CSOH

Answer 1

A) at initial point

HN3 + H2O <------> N3^- + H3O⁺

Ka= [H3O⁺] [N3^- /[HN3] = 1.9×10^-5

at equillibrium ,

[HN3] = 0.100 -x

[N3^-] = x

[H3O+] = x

x²/(0.100 -x) = 1.9×10^-5

solving for x

x = 0.001369

[H3O+] = 0.001369M

pH = -log(H3O+]

pH = -log(0.001369)

pH = 2.86

B) 12.50 ml CsOH addition

The reaction between CsOH and HN3 is 1:1 molar reaction

HN3 + OH^- -----> H2O + N3^-

no of moles of HN3 = (0.100mol/1000ml)×50ml = 0.0050

0.0050 moles of HN3 react with 0.0050moles of CsOH

Volume CsOH solution containing 0.0050moles of OH^- = (1000ml/0.200mol)×0.0050mol = 25ml

Therefore,

25 ml addition of CsOH is half equivalence point

at half equivalence point

pH = pKa

pKa = -log(Ka)

pKa = - log(1.9×10^-5 )

pKa = 4.72

Therefore,

pH = 4.72

iii) 50ml CsOH addition

No of moles of OH^- excess added = ( 0.200mol/1000ml)×25ml = 0.005

Total volume = 50ml + 50ml =100ml

[OH^-] = (0.005mol/100ml)×1000ml = 0.05M

pOH = -log[OH^- ]

pOH = -log(0.05)

pOH = 1.30

pH = 14- pOH

pH = 14 -1.30

pH = 12.70

D) 60ml CsOH addition

excess moles of CaOH added = (0.200mol/1000ml)× 35ml = 0.007mol

Total volume = 110ml

[OH^- ]= (0.007mol/110ml)×1000ml = 0.0636M

pOH = - log (0.0636) = 1.20

pH = 14 - 1.20

pH = 12.80