Pseudo approach : At `25^(@)C` and at a constant pH of 5, the hydrolysis of reactant, A, proceeds with a constant half-life of 500 min. At this temperature, but at a pH of 4, the half life ios constant at 50 min. Find the rate law.
Strategy : `C_(H^(+))= 10^(-pH)mol//L`, since half life `(t_(1//2)prop 1//rate constant)` changes with change of pH, it impiles that reaction rate also depends upon `[H^(+)]`. Thus, the rate law should of the following type
`Rate=K[A]^(x)[H^(+)]^(y)`
To find this rate law, we need to determine two unknowns : x and y. We can simplify the situation by pseudo approach which allows us to find x and y stepwise.

Explanation

There are two experiments, each of which is carried out at constant pH, i.e, at constant `[H^(+)]`. Thus, in in any given experiment, we can cosider the rate law to be of the type
`Rate = k'[A]^(x)`
where `k' = k[H^(+)]^(y)`. This modified rater law can be used as long as the pH (or concentration of `H^(+)` ions) is constant. At constant pH, reaction proceeds with a constant half-life. Since half-life is constant only for a first order reaction, we conclude that `x = 1` i.e., reaction is first order with respect to reactant 'A' and the rate constant k' is a first order rate constant. k' is related to half-life by the relation
`k' = (0.693)/(t_(1//2))`
To find out the value of exponent 'y', we must change the pH. When we change the pH from 5 to 4, the conc. of `H^(+)` ions changes from `10^(-5) mol L^(-1)` , i.e., there is ten times increase in the concentration of `H^(+)` ions. Since a ten fold increase in `[H^(+)]` decreases the half-life ten times (from 500 to 50), the value of k' increases ten times.
`underset("increase")underset("ten fold")underset(uarr)(k') = underset("increase")underset("ten fold")underset(uarr)(K[H^(+)]^(y))`
Ths can be possible only if `y = 1`, i.e., the reaction is also first order with respect to `H^(+)`. Thus, the complete rate law is
`Rate = k[A][H^(+)]`