what is the PH of a solution prepared by adding enough water to 15.00g of NaNO2 to make 275ml of solution

Answer 1

Calculate the molarity.
M = mols/L and mols = grams/molar mass
I get 15.00/69 = 0.217 mols
and M = 0.217/0.275 L = 0.790 M. You need to confirm this. I just estimated the molar mass.
The pH is determined by the hydrolysis (reaction with water) of the nitrite ion.

NO2^- + HOH ==> HNO2 + OH^-

Kb = Kw/Ka = (HNO2)(OH^-)/(NO2^-) = ??

You know Kw and ka for HNO2 so you can calculate Kb.
Now wet up an ICE table.

I = Initial concentration:
C = change in concentration:
E = equilibrium concentration.

Initial:
(NO2^-) = 0.790 M
(HNO2) = 0
(OH^-) = 0

change:
(OH^-) = +y
(HNO2) = +y
(NO2^-) = -y

Equilibrium:
(NO2^-) = 0.790 - y
(HNO2) = 0 + y = y
(OH^-) = 0 + y = y

Now substitute the equilibrium values above into the Kb expression above, and solve for y. That will be equal to the (OH^-), change that to pOH, then subtract from 14 to obtain pH.
Post your work if you get stuck.

Answer 2

thank you so much!