The degree of hydrolysis of 02.M solution of NaHCO₃ = 2.33 * 10⁻⁸
Given data :
Concentration of NaHCO₃ = 0.2 M
Ka1 = 4.3 * 10⁻⁷
Ka2 = 4.8 * 10⁻¹¹
Determine the degree of hydrolysis
NaHC0₃ ----> Na⁺ + HCO₃
where : Na⁺ is a stronger and HCO₃ is weaker
Therefore;
hydrolysis will occur at HCO₃
HCO₃ + excess water --> H₂CO₃ + OH⁻ --- kh
kh = hydrolysis constant
Also
Kh = Kw / Ka1
where ; Kw = ionic product of water ( 10⁻¹⁴ ) , Ka1 = dissociation constant
Finally the degree of hydrolysis is
Kh = ( 10⁻¹⁴ ) / 4.3 * 10⁻⁷
= 2.33 * 10⁻⁸
Hence we can conclude that The degree of hydrolysis of 02.M solution of NaHCO₃ = 2.33 * 10⁻⁸
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