हिंदी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान कक्षा ११

The pH of 0.005M codeine (C18H21NO3) solution is 9.95. Calculate its ionization constant and pKb.

Advertisements
Advertisements

प्रश्न

The pH of 0.005M codeine (C18H21NO3) solution is 9.95. Calculate its ionization constant and pKb.

संख्यात्मक
Advertisements

उत्तर

c = 0.005

pH = 9.95

pOH = 4.05

pH = – log (4.105)

`4.05 = -log["OH"^-]`

`["OH"^-] = 8.91 xx 10^(-5)`

`"c"  alpha = 8.91 xx 10^(-5)`

`alpha = (8.91 xx 10^(-5))/(5xx10^(-3)) = 1.782 xx 10^(-2)`

Thus `"K"_"b" = "c"  alpha^2`

`= 0.005 xx (1.782)^2 xx 10^(-4)`

`= 0.005 xx 3.1755 xx 10^(-4)`

`= 0.0158 xx 10^(-4)`

`"K"_"b" = 1.58 xx 10^(-6)`

`"Pk"_"b" = - log "K"_"b"`

`= - log (1.58 xx 10^(-6))`

= 5.80

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 6: Equilibrium - EXERCISES [पृष्ठ २३७]

APPEARS IN

एनसीईआरटी Chemistry - Part 1 and 2 [English] Class 11
अध्याय 6 Equilibrium
EXERCISES | Q 7.51 | पृष्ठ २३७
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×