Calculate the pH of 0.1 MNH_(3) solution. K_(b) = 1.8 ** 10^(-5)
What is the pH of 0.01 M glycine solution? [For glycine, K_(a_(1))=4.5xx10^(-3) and K_(a_(2))=1.7xx10^(-10) at 298 K ]
Calculate pH of a resultant solution of 0.1 M HA (K_(a)=10^(-6)) and 0.5 M HB (K_(a)=2xx10^(-6)) at 25^(@)C.
Find the pH of " 0.1 M NaHCO_(3)" . Use data (K_(1)=4xx10^(-7),K_(2)=4xx10^(-11) for H_(2)CO_(3), log 4=0.6) :
STATEMENT-1: When 0.1 M weak diprotic acid H_(2)A dissociated with its dissociation constants K_(a_1)=10^(-3) and K_(a_2)=10^(-8) , then [A^(2-)] is almost equal to 10^(-3) M STATEMENT-2: Since K_(a_2)ltltK_(a_1) for 0.1 M H_(2)A,so[A^(2-)] is negligible w.r.t. [HA^(-)]
pH of a 0.01 M solution (K_(a)=6.6xx10^(-4))
0.1 M H_(2)S has K_(1)=10^(-5) & K_(2)=1.5xx10^(-12) . What will be the concentration of S^(-2) in the solution.
Ratio of [HA^(+)] in 1L of 0.1M H_(3)A solution [K_(a_(1))=10^(-5),K_(a_(2))=10^(-8)K_(a_(3))=10^(-11)] & upon addition of 0.1 mole HCl to it will be :
[H^(+)] concentration in 0.01 M H_2O_2 solution (K_(a_(1))=3xx10^(-12) and K_(a_2)~~0) is xxM .Fill first two digits of 10^(8)x as answer.
Selenious acid (H_(2)SeO_(3)) , a diprotic acid has K_(a1)=3.0xx10^(-3) and K_(a2)=5.0xx10^(-8) . What is the [OH^(-)] of a 0.30 M solution of selenious acid?