The equilibrium constant for the ionisation of `RNH_(2(g))` in water as `RNH_(2(g)) + H_(2)O_((l)) hArr "RNH"_(3(aq))^(+) + OH_((aq))^(-) ` is `8 xx 10^(-6) ` at `25^(@)` C . Find the pH of a solution at equilibrium when pressure of `RNH_(2(g)) ` is 0.5 bar.
A
`~~12.3`
B
`~~11.3`
C
`~~11.45`
D
None
Text Solution
Verified by Experts
The correct Answer is:
B
Topper's Solved these Questions
CHEMICAL EQUILIBRIUM
NARENDRA AWASTHI|Exercise Level 1 (Q.93 To Q.122)|1 Videos
CHEMICAL EQUILIBRIUM
NARENDRA AWASTHI|Exercise Level 2|1 Videos
ATOMIC STUCTURE
NARENDRA AWASTHI|Exercise level 2|1 Videos
DILUTE SOLUTION
NARENDRA AWASTHI|Exercise Level 3 - Match The Column|1 Videos
Similar Questions
Explore conceptually related problems
Calculate enthalpy of ionisation of OH^(-) ion. Given: H_(2)O_((l)) to H_((aq))^(+) + OH_((aq))^(-) , DeltaH^(@) = -285.83 KJ
For the reaction 2H_(2)O_((l)) rarr H_(3)O_((aq))^(+) + OH_((aq))^(-) , the value of Delta H is
For Ag^(+)(aq) + 2NH_(3(aq)) harr Ag(NH_(3))_(2(aq))^(+) k = 1.7 xx 10^(7) at 25^(@)C , for this equilibrium state, which is correct?
The equilibrium constant for the reaction , H_(2) (g) + CO_(2) (g) hArr H_(2) O (g) + CO (g) is 16 at 1000^(@) C . If 1.0 mole of H_(2) and 1.0 mole of CO_(2) are placed in one litre flask , the final equilibrium concentration of CO at 1000^(@) C is
The equilibrium constant for the reaction N_(2(g)) +O_(2(g)) hArr 2NO_((g)) is 4 xx 10^(-4) at 200K. In presence of a catalyste, equilibrium is attaincd ten times faster. Therefore, the equilibrium constant in the presence of the catalyst at 200K is
Write expression for the equilibrium constant, Kc for each of the following reactions. CH_(3)COOC_(2)H_(5(aq))+H_(2)O(l) hArr CH_(3)COOH_((aq))+C_(2)H_(5)OH_((aq))
Passage: When all the coefficients in a balanced chemical equation are multiplied by a constant factor X the equilibrium constant (originally K) becomes K^J . Similarly, when balanced equations are added together, the equilibrium constant for the combined process is equal to the product of the equilibrium constants for each step. Equilibrium constant of the reversed reaction is numerically equal to the reciprocal of the equilibrium constant of the original equation. Unit of K_p = ("atm")^(Deltan) , Unit of K_c =("mol" L^(-1))^(Deltan) The equilibrium constants for the following reactions at 1400 K are given: 2H_2O_((g)) harr 2H_(2(g))+O_(2(g)) , K_1=2.1 xx 10^(-13) 2CO_(2(g)) harr 2CO_((g))+O_(2(g)) , K_2=1.4 xx 10^(-12) Then the equilibrium constant K for the reaction, H_(2(g))+CO_(2(g)) harr CO_((g)) + H_2O_((g)) is
The equilibrium constant for the reaction: N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g) at 725K is 6.0xx10^(-2) . At equilibrium [H_(2)]=0.25 "mol" L^(-1) and [NO_(3)]=0.06 "mol" L^(-1) Calculate the equilirbium concentration of N_(2) .