Question

14. What is pH of 0.0002 M HI solution? What will be pH if 1 mL of the above was added to 999 mL of pure water?

15. What is pH of mixture prepared by mixing 20 mL 0.07 M NaOH and 13 mL 0.09 M HCl?

16. Calculate pH of 1 M solution of nitrous acid, pK_a=3.37.

17. Calculate pH of 0.001 M solution of benzoic acid, pK_a=4.19.

Answer 1

14)

i) HI is strong acid and completely dissociates

HI - - - - - > H+ + I^-

[HI] = [H+] = 0.0002M

pH = - log[H+]

= - log(0.0002)

= 3.70

ii) when 1ml of 0.0002M HI solution is added to 99ml of water, the dilution factor is 100

therefore,

[H+] = 0.0002M/100 = 0.000002M

pH = - log(0.000002)

= 5.70

15)

NaOH + HCl - - - - > NaCl + H2O

1:1 molar reaction

No of moles of HCl =( 0.09mol/1000ml)×13ml = 0.00117

No of moles of NaOH = ( 0.07mol/1000ml)×20ml = 0.0014

0.00117mole of HCl react with 0.00117 mole of NaOH

Remaining mole of NaOH = 0.0014 - 0.00117 = 0.00023

Total Volume = 20ml + 13ml = 33ml

NaOH is strong base and completely dissociates

[NaOH] = [OH-] = (0.00023/33ml)×1000ml = 0.0070M

pOH = - log[OH-]

   = - log(0.0070)

   = 2.15

pH + pOH = 14

   pH = 14 - pOH

   = 14 - 2.15

   = 11.85

16)

pKa = - logKa

- logKa = 3.37

   Ka = 4.27×10^-4

HNO2 <-------> H+ + NO2^-

   Ka = [H⁺] [NO2^-] /[HNO2] = 4.27×10^-4

at equillibrium

[HNO2] = 1-x

[H+] = x

[NO2-] = x

So,

x²/1-x = 4.27×10^-4

Solving for x

  x = 0.02045

[H+] = 0.02045M

pH = - log(0.02045)

   =1.69

17)

pKa = 4.19

   Ka = 6.46×10^-5

dissociation of Benzoic acid is

   C6H5COOH - - - - > C6H5COO- + H+

   Ka = [C6H5COO-] [H+] /[C6H5COOH] = 6.46×10^-5

  x²/0.001 - x = 6.46×10^-5

solving for x

   x = 0.0002239

[H+] = 0.0002239

pH = - log(0.0002239)

   = 3.65