1.

Calculate the hydrolysis constant (ii) degree of hydrolysis (iii) pH of 0.05 M sodium carbonates solution pK_a for HCO_3^- is 10.26.

Answer»

Solution :`h=SQRT((K_w)/(K_a timesc))`
GIVEN `K_w=1 times10^-14`
`c=0.05M`
`pK_a=10.26`
`pK_a=-logK_a`
`K_a=antilog of (-pK_a)`
`K_a=antilog of (-10.26)`
`K_a=5.49 times10^-11`
`h=sqrt((1 times10^-14)/(5.49 times10^-11 times0.05))=sqrt(3.642 times10^-3)`
`h=6.034 times10^-2`
(ii) Degree of HYDROLYSIS `K_h=K_w/K_a= (1 times10^-14)/(5.49 times10^-11)=1.82times10^-4`
(III) `pH=7+(pK_a)/2+logC/2`
`=7+10.26/2+(log(0.05))/2=7+5.13+((-1.30)/2)=7+5.13-0.65`
pH=11.48


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