1.

0.1 M HQ acid has pH=3 then findits ionization constant.

Answer»

`3xx10^(-1)`
`1xx10^(-3)`
`1xx10^(-5)`
`1xx10^(-7)`

Solution :`{:(,HQoversettolarr, H^+ +, Q^-),("Initial " , 0.1 M, O,O),("Ionized MOLE",X,x,x),(,darr,darr,darr),("Mole at EQUILIBRIUM",0.1-x,x,x):}`
pH=3
`THEREFORE [H^+]=10^(-3)`
`therefore x=10^(-3)`
`therefore K_a=((x)xx(x))/(0.1-x)`
`=(10^(-3))^2/(0.1-10^(-3)) approx 10^(-6)/0.1 = 1xx10^(-5)`


Discussion

No Comment Found