If a soup has pH 6.7. What would be its taste?

Explanation

(a) Let rate `= k[CH_(3)COF]^(a) [H_(2)O]^(b)`
Uisng Ostwald isolation methof,
In first case: `[H_(2)O] gt gt [CH_(3)COF]_(0)`
In second case: `[H_(2)O] lt lt [CH_(3)COF]_(0)`
In first case, determine the order of reaction w.r.t. `(CH_(3)COF)`, Since `[H_(2)O]` is very high.
The reaction is not of zero order as the rate of reaction changes with time. Uisng first order reaction we find:
`|{:(t (min),[CH_(3)COF] (M),k[H_(2)O]^(b) (min^(-1)) = (2.3)/(t)log.([CH_(3)COF]_(0))/([CH_(3)COF]_(t))),(0,0.01000,-),(10,0.00857,0.0154),(20,0.00735,0.0154),(40,0.00540,0.0154):}|`
`:. k[H_(2)O]^(b) = 0.0154 min^(-1)`
Hence, order of reaction w.r.t. `(CH_(3)COF) = 1`
(b) Now the order of reaction w.r.t. `H_(2)O` is determined by uisng first order reaction.
`|{:(t//min,[H_(2)O] (M),k[CH_(3)COF]^(a) (min^(-1)) = (2.3)/(t) log.([H_(2)O]_(0))/([H_(2)O])),(0,0.0200,-),(10,0.0176,0.0128),(20,0.0156,0.0124),(40,0.0122,0.0124):}|`
Average `= 0.0125 min^(-1)`
Hence, the reaction is first order w.r.t. `(H_(2)O)`.
`:. k[CH_(3)COF]^(1) = 0.0125 min^(-1)`
`k` in case II `= 0.0125//0.800 M`
`= 0.0156 M^(-1) min^(-1)`
`k` in case I
`k[H_(2)O]^(1) = 0.0154 min^(-1)`
`k = 0.0154//1.00 M = 0.0154 M^(-1) min^(-1)`