The `[H^(+)]` in `0.2M` solution of formic acid is `6.4xx10^(-3)` mol `litre^(-1)`. To this solution formate is added so as to adjusrt the conc.of sodium formate to one mol per litre. What will be pH of this solution ? `(K_(a) for HCOOH=2.4xx10^(-4))` and degree of dissociation of `HCOONa = 0.75`)

The concentration of H^(+) ion in a 0.2 M solution of HCOOH is 6.4xx10^(-3) "mole" L^(-1) . To this solution HCOONa is added so as to adjust the concentration of HCOONa to one mole per litre. What will be the pH of this solution? K_(a) for HCOOH is 2.4xx10^(-4) and the degree of dissociation of HCOONa is 0.75

The concentration of hydrogen ions in a 0.2M solution of formic acid is 6.4 xx 10^(-3) mol L^(-1) . To this solution, sodium formate is added so as to adjust the concentration of sodium formate to 1 mol L^(-1) . What will be the pH of this solution? The dissociation constant of formic acid is 2.4 xx 10^(-4) and the degree of dissociation fo sodium formate is 0.75 .

The concentration of hydrogen ions in a 0.2M solution of formic acid is 6.4 xx 10^(-3) mol L^(-1) . To this solution, sodium formate is added so as to adjust the concentration of sodium formate to 1 mol L^(-1) . What will be the pH of this solution? The dissociation constant of formic acid is 2.4 xx 10^(-4) and the degree of dissociation fo sodium formate is 0.75 .

What is the pH of a 0.05 M solution of formic acid? (K_(a)=1.8xx10^(-4))

If K_(a) for formic acid is 2.0 xx 10^(-4) mol L^(-1) then K_(h) for HCOO^(-) is

Calculate the amount of HCOONa that is to be added to 0.05 (M) 1 L HCOOH to make the overall pH of the solution 4.00 (Given : K_(a) of HCOOH = 2.4 xx 10^(-4) ).

What would be the pH of an aqueous solution of sulphuric acid which is 5xx10^(-5) mol "litre"^(-1) in concentration.