......................HONH2 + HOH ==> HONH3^+ + OH^-
I.......................0.0991........................0................0
C.........................-x............................x..................x
E......................0,0991-x.....................x..................x
Plug the E line into the Kb expression for HONH2 and solve for x = (OH^-).
I would then convert that to pOH and for part b use pH + pOH = pKw = 14. YOu know Kw and pOH, solve for pH.
Post your work if you get stuck.
A)Calculate [OH-] of a 9.91×10-2 M aqueous solution of hydroxylamine (HONH2, Kb = 1.1×10-8).
b)Calculate the pH of the above solution.
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1 answer