Asked by Anonymous
What is the original molarity of a solution of ammonia whose pH is 10.83
Answers
Answered by
DrBob222
First, pH + pOH = pKw = 14. You know pH and pKw, solve for pOH, then
pOH = -log(OH^-). Then,
Kb = (NH4^+)(OH^-)/(NH3) and solve for (NH3)
Post your work if you get stuck.
pOH = -log(OH^-). Then,
Kb = (NH4^+)(OH^-)/(NH3) and solve for (NH3)
Post your work if you get stuck.
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