Correct Answer - A
`{:(,HCOOH(aq.)hArrH^(+)(aq.)+HCOO^(-)(aq.)),("Initial (M)":," 0.100 0.000 0.000"),("Change (M)":," -x +x +x"),("Equilibrium (M)":,bar(" (0.100-x) x x ")):}`
`K_(a)=(C_(H)C_(HCOO^(-)))/(C_(HCOOH))`
`1.7xx10^(-4)=(x^(2))/(0.100-x)`
This equation can be rewritten as
`x^(2)+1.7xx10^(-4)x-1.7xx10^(-5)=0`
which fits the quardatic equation `ax^(2)+bx+c=0`. We can often apply a simpliffying approximation to this type of problem. Since HCOOH is a weak acid, the extent of ionization must be small. Therefore, x is small compared with `0.100`. As a general rule, if the quantity x that is subtracted form the original concentration of the acid (`0.100` M in this case) is equal to or less than `5%` of the original concentration. we can assume `0.100-x~~0.100`. This removes x form the denominator of the equilibrium constant expression. If x is more than `5%` of the original value, then we must solve the quadratic equation. Normally, we can apply the approximation in case where `K_(a)` is small (equal to or less than `1xx10^(-4))` and the initial concentration of the acid is high (equal to or greater than `0.1` M). In case of doubt, we can always solve for x by the approximation methodand then check the the validity of our apporiximation. Assuming that `0.100-x~~0.100`, we have
`(x^(2))/(0.100)=1.7xx10^(-4)`
`x^(2)=1.7xx10^(-5)`
Taking the square root of both sides, we obtain
`x=4.1xx10^(-3)M`
To check the validity of our appoximation
`(0.0041M)/(0.100M)xx100%=4.1%`
This show that the quantity x is less than `5%` of the original concentration of the acid. Thus, our approximation was justified. Note that we omitted the contribution to `C_(H^(+))` by water. Except in very dilute acid solution, this omission is always valid, because the `C_(H^(+))` due to water is negligible (small) compared with that due to an acid.
