What is the concentration of `CH_(3)COOH(aq)` in a solution prepared by dissolving `0.01` mole of `NH_(4)^(+)CH_(3)COO^(-)` in `1L H_(2)O`? Report your answer after multiplying by `9xx10^(5)`.

What is the concentration of CH_(3)COOH(aq.) in a solution prepared by dissolving 0.01 mole of NH_(4)^(+)CH_(3)COO^(-) in 1 L H_(2)O ? [K_(a) (CH_(3)COOH))=1.8xx10^(-5)),K_(b) (NH_(4)OH)=1.8xx10^(-5))]

NH_(4)OH , NaOH, CH_(3)COOH , H_(2)S

The concentraion of CH_(3)COOH and HCN is equal. Their pH 3.0 and 2.0 respectively. If K_(a) of CH_(3)COOH is 1.8 xx 10^(-5) then K_(a) value of HCN is :

What is the concentration of CH_(3)COOH which can be added to 0.5M HCOOH solution so that dissociation of both is same. K_(CH_(3)COOH) = 1.8 xx 10^(-5), K_(HCOOH) = 2.4 xx 10^(-4)

Calculate degree of hydrolysis(h) and pH of solution obtained by dissolving 0.1 moles of CH_(3)COOHNa in water to get 100L of solution .Take K_(a) of acetic acid =2xx10^(-5) at 25^(@)C .

The pH of solution, containing 0.1N HCl and 0.1N CH_(3)COOH(K_(a)=2xx10^(-5)) is

Caculate [H^(+)] in a solution that is 0.1 M both in CH_(3)COOHandCH_(3)COONa (pK_(a)CH_(3)COOH=4.7) :-

What is the [OH^(-)] concentration of a 0.04 M solution of CH_(3)COONa? [K_(a)" of "CH_(3)COOH=2xx10^(-5),log2=.30]

Calculate the hydrogen ion concentration ( "in mol"//dm^3 ) in a solution containing 0.04 mole of acetic acid and 0.05 mole of sodium acetate in 500 mL of solution.Dissociation constant for acetic acid is 1.75xx10^(-6) Report your answer after multiplying by 2xx10^(6)

Calculate the pH of a solution containing 0.1 M CH_(3)COOH and 0.15 M CH_(3)COO^(-) . (K_(a) "of" CH_(3)COOH=1.8xx10^(-5))