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:

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

∴ C H2SO3 (aq) = 0.163 M

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

∴ 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

it have destructive character

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.