#"Hydroxylamine"#, #HONH_2# is a weak Bronsted base with #K_b=10^-8#. What is #pH# for a #0.024*mol*L^-1# solution?

1 Answer
Apr 21, 2017

Answer:

We need the stoichiometric equation............and get #pH~=9.#

Explanation:

#HONH_2(aq) + H_2O(l) rightleftharpoonsHONH_3^+ + HO^-#

And then we write the equilibrium expression:

#K_b=10^-8=([HONH_3^+][HO^-])/([HONH_2])#,

And we make the usual approximations, and invoke #x# as the moles of hydroxylamine that associate,

#K_b=x^2/(0.024-x)#

And if #"0.024>>>"x#, then #x~=sqrt(K_bxx0.024)#

i.e. #x~=sqrt(10^-8xx0.024)#

#x_1=1.55xx10^-5*mol*L^-1#, which is indeed small compared to #0.024#, but let us recycle that value back into the equation......

#x_2=1.55xx10^-5*mol*L^-1#,

So, from the equilibrium expression, we now know #x=[HO^-]=1.55xx10^-5*mol*L^-1#.

How does this help us, as we were asked to measure #pH#?

Well in aqueous solution under standard condtions, #pH+pOH=14#, thus #pH=14-pOH=14+log_10{1.55xx10^-5}~=9#. This is slightly basic, as we would expect for a weak Bronsted base.