Home
Class 12
CHEMISTRY
A 0.025 m solution of monobasic acids ha...

A `0.025 m` solution of monobasic acids has a freezing point of `-0.060^(@)C`. What are `K_(a)` and `pK_(a)` of the acid? `(K_(f)=1.86^(@)C)`

Text Solution

Verified by Experts

`DeltaT_(f)=iK_(f)m` , `0.60 =ixx1.86xx0.25`
`:.i=1.29`

`:.i=1-alpha+alpha=1+alpha`
`:.alpha=0.29`
`K_(a)=(Calpha)^(2)/(1-alpha)=(0.025xx(0.29^(2)))/(1-0.29) =3xx10^(-3) rArr pK_(a)=2.5`
Promotional Banner

Similar Questions

Explore conceptually related problems

The freezing point of a 0.05 molal solution of a non-eletrolyte in water is ( K_(f)=1.86Km^(-1) )

A certain solution of 1 m benzoic acid in benzene has a freezing point of 3.1^(@)C and a normal boiling point of 82.6 ^(@)C . The freezing point of benzene is 5.5^(@)C . And its boiling point is 80.1^(@)C . Analyze the state of the solute (benzoic acid ) at two temperature and comment .

The freezing point of a 0.05 m BaCl_(2) in water ( 100% ionisation) is about (K_(f)=1.86 Km^(-1)) :

If 10.0 g of a non-electrolyte dissolved in 100 g of water lowers the freezing point of water by 1.86^(@)C , the molar mass of the non-electrolyte is ( K_(f)=1.86 Km^(-1) )

AN aqueous solution of a substacne freezes at -0.186^(@)C. The boiling point of the same solution (k _(f) =1.86 Km^(-1), K_(b)=0.512 Km^(-1)) is :

A solution of urea in water has boiling point of 100.15^(@)C . Calculate the freezing point of the same solution if K_(f) and K_(b) for water are 1.87 K kg mol^(-1) and 0.52 K kg mol^(-1) , respectively.

A solution containing 25.6 g of sulphur, dissolved in 1000 g of naphthalene whose melting point is 80.1^(@)C gave a freezing point lowering of 0.680^(@)C . Calculate the formula of sulphur ( K_(f) for napthalene = 6.8 K m^(-1) )

A solution of 1.25 g of 'P' in 50 g of water lowers freezing point by 0.3^(@) C . Molar mass of 'P' is 94. K_("(water)") = 1.86 K kg mol^(-1) . The degree of association of ‘P’in water is