Question

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

Answer 1

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.