Question

The ionization constants of HF, HCOOH and HCN at 298 K are $6.8 \times 10^{-4}$, $1.8 \times 10^{-4}$ and $4. 8 \times 10^{-9}$ respectively. Calculate the ionization constant of the corresponding conjugate base.

Answer 1

"${ K }_{ a }, { K }_{ b } \quad and \quad { K }_{ w }$ are dissociation constants of weak acid, conjugate base, and Ionic product of water respectively.

At the temperature, 298 K

${ K }_{ w }\quad =\quad 1.0\quad \times \quad { 10 }^{ -14 }$

${ K }_{ a }.{ K }_{ b }\quad =\quad { K }_{ w }$

$HF/{ F }^{ -2 }$

${ K }_{ HF }\quad \times \quad { K }_{ F }\quad =\quad 1.0\quad \times \quad { 10 }^{ -14 }$

$6.8\quad \times \quad 1{ 0 }^{ -4 }\quad \times \quad { K }_{ { F }^{ - } }\quad =\quad 1.0\quad \times \quad 1{ 0 }^{ -14 }$

${ K }_{ { F }^{ - }\quad  }\quad =\quad \frac { 1.0\quad \times \quad 1{ 0 }^{ -14 } }{ 6.8\quad \times \quad 1{ 0 }^{ -4 } } \quad =\quad 1.5\quad \times \quad 1{ 0 }^{ -4 }$

$HCOOH/HCO{ O }^{ - }$

${ K }_{ HCOOH }\quad \times \quad { K }_{ HC{ oo }^{ - } }\quad =\quad 1.0\quad \times \quad { 10 }^{ -14 }$

$1.8\quad \times \quad { 10 }^{ -4 }\quad \times \quad { K }_{ HC{ oo }^{ - } }\quad =\quad 1.0\quad \times \quad { 10 }^{ -14 }$

${ K }_{ HC{ oo }^{ - } }\quad =\quad 1.0\quad \times \quad { 10 }^{ -14 }$

$Total\quad =\quad 5.6\quad \times \quad { 10 }^{ -11 }$

$HCN/C{ N }^{ - }$

${ K }_{ HCN }\quad \times \quad { K }_{ { CN }^{ - } }\quad =\quad 1.0\quad \times \quad { 10 }^{ -14 }$

$4.8\quad \times \quad { 10 }^{ -9 }\quad \times \quad { K }_{ { CN }^{ - } }\quad =\quad 1.0\quad \times \quad { 10 }^{ -14 }$

${ K }_{ { CN }^{ - } }\quad =\quad \frac { 1.0\quad \times \quad { 10 }^{ -14 } }{ 4.8\quad \times \quad { 10 }^{ -9 } }$

$Total\quad =\quad 2.08\quad \times \quad { 10 }^{ -6 }$"

Answer 2

HEY MATE......☺✌☺

THE ANSWER IS $\huge{28×10^-8}$