1. Calculate the approximate [OH-] and [NH4+] in a 0.40 M ammonia solution, NH3(aq). NH3(aq) +...
1. Calculate the approximate [OH-] and
[NH4+] in a 0.40 M ammonia solution,
NH3(aq).
NH3(aq) + H2O(l) ↔ OH-(aq) +
NH4+(aq). Kb = 1.75 x
10-5M.
2. Calculate the pH of 0.178 M ammonia.
NH3(aq) + H2O(l) ↔ OH-(aq) +
NH4+(aq) Kb = 1.75 x
10-5
Homework Answers
1)
NH3 + H2O ----> OH- + NH4 + , Kb = 1.75*10^-5
0.40 0 0 (at t=0)
0.40-x x x (at equilibrium)
Kb = x*x / 0.40-x
1.75*10^-5 = x^2 / 0.40-x
0.000007 - 0.0000175x = x^2
x^2 + 0.0000175x - 0.000007 = 0
Solving above quadratic equation,
x = 0.00264
[OH-] = [NH4+] = x = 0.00264 M ..... Answer
2) Similarly as above,
Kb = x^2 / 0.178-x
0.0000175 * (0.178-x) = x^2
0.000003115 - 0.0000175x = x^2
x^2 + 0.0000175x - 0.000003115 = 0
x = 0.00176
[OH-] = [NH4 +] = x = 0.00176 M .... Answer
Let me know in case of any doubts.
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1. Calculate the approximate [OH-] and
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