1.7*102 and k2=6.4*108 at 25.0°c. calculate the ph of a issociation constants of sulfurous acid (h2so3) are kal solution of sulfurous acid. at 25.0°c. calculate the ph of a 0.163m aqueous

pH = 1.348

Explanation:

H2SO3 + H20 ↔ H3O+  +  HSO3-

∴ Ka1 = 1.7 E-2 = ( [ H3O+ ] * [ HSO3- ] ) / [ H2SO3 ]

∴ C H2SO3 (aq) = 0.163 M

HSO3-  + H2O ↔ H3O+  +  SO32-

∴ Ka2 = 6.4 E-8 = ( [ H3O+ ] * [ SO32- ] ) / [ HSO3- ]

2H2O ↔ H3O+  +  OH-

∴ Kw = 1 E-14 = [ H3O+ ] * [ OH- ]

since Ka1 >>> Ka2 ( difference between the constants is greater than the order of E3 );   then the most importan equilibrium is Ka1 and we can assume that [ SO32- ] <<< [ HSO3- ]

mass balance:

⇒ C H2SO3 = 0.163 = [ H2SO3 ] + [ HSO3- ](1)

charge balance:

⇒ [ H3O+ ] = [ HSO3- ](2)  where [ OH- ] and [ SO32- ] is neglected. the first comes from water and the second comes from Ka2.

(2) in (1):

⇒ [ H2SO3 ] = 0.163 - [ H3O+ ](3)

(2) and (3) in Ka1:

⇒ 1.7 E-2 = [ H3O+ ]² / ( 0.163 - [ H3O+ ] )

⇒ [ H3O+ ]² + 1.7 E-2 [ H3O+ ] - 2.771 E-3 = 0

⇒[ H3O+ ] = 0.045 M

⇒ pH = - Log [ H3O+ ] = - Log ( 0.045 )

⇒ pH = 1.348

What is the question

☆ionisation energy increases across a period from na -> ar☆ionisation energy decreases down a group from li -> fr☆ionic radius decreases across a period from na -> cl☆ionic radius increases down a groupfrom f -> ii hope it clarifies your doubts