A 0.1M solution of weak acid HA is 1% dissociated at 298k . what is its K_(a) ? what will be the new degree of dissociation of HA and pH when 0.2M of NaA is added to it.
Calculate the pH of 0.10 M solution of NH_(4)Cl. The dissociation constant (K_(b)) of NH_(3) is 1.8 X 10^(-5)
Calculate the pH of 0.10 M solution of NH_4CI . The dissociation constant (K_b) "of" NH_3 "is" 1.8 xx 10^(-5) .
The pH of a 0.1M solution of NH_(4)OH (having dissociation constant K_(b) = 1.0 xx 10^(-5)) is equal to
Calculate the degree of hydrolysis of 0.1 M solution of sodium acetate at 298 K : K_(a) = 1.8 xx 10^(-5) .
The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 ohm. What is the cell constant and molar conducti vity of 0.001 M KCI solution, if the conductivity of this solution is 0.146xx10^(-3)"ohm"^(-1)cm^(-1) at 298 K?
0.01 M monoprotonic acid is 10% ionised, what is K_(a) of the solution ?
What will be the pH at the equivalence point during the titration of a 100 mL 0.2 M solution of CH_(3)CCOONa with 0.2 M solution of HCl ? K_(a)=2xx10^(-5)
Total number of solutions from the following which has pHlt7 at 25^(@)C . (K_(a)(CH_(3)COOH)=1.8xx10^(-5),K_(b)(NH_(4)OH)=1.8xx10^(-5)) (i) 10^(-8) MHCl (ii) 0.01 M solution of NH_(4)Cl (iii) 0.01 M solution of CH_(3)COONa (iv) 0.01M solution is B(OH)_(3) (v) 0.01M solution of CH_(3)COONH_(4)
The conductivity of 0.10 M solution of KCl at 298 K is 0.025 S cm Calculate its molar conductivity.
