Asked by Addy
Refer to Equilibrium Constants. Calculate the pH of a solution resulting from the addition of 50.0 mL of 0.30 M HNO3 to 50.0 mL of 0.30 M NH3.
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Answered by
Anonymous
After addition, all of the NH3 becomes NH4^+ and the pH is determined by dissociation of H^+.
NH4^+ -----------------> NH3 + H^+
Calculate new molarity after addition of acid.
Molarity of NH4^+=(0.3M*0.050L)/(0.100L)=0.15M
I......0.15 M...................0............0
C......-X.............,.......+X..........+X
E.....0.15 M-X..............X...........X
Kw/Kb=Ka=[products]/[reactants]
Ka=[x][x]/[0.15M-x]
5% dissociation allows for simplification....
Ka=x^2/[0.15M]
X=√(Ka*0.15M)
pH=-log[x]
Answer with the correct number of significant figures and.check 5% dissociation.
NH4^+ -----------------> NH3 + H^+
Calculate new molarity after addition of acid.
Molarity of NH4^+=(0.3M*0.050L)/(0.100L)=0.15M
I......0.15 M...................0............0
C......-X.............,.......+X..........+X
E.....0.15 M-X..............X...........X
Kw/Kb=Ka=[products]/[reactants]
Ka=[x][x]/[0.15M-x]
5% dissociation allows for simplification....
Ka=x^2/[0.15M]
X=√(Ka*0.15M)
pH=-log[x]
Answer with the correct number of significant figures and.check 5% dissociation.
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