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. 0.5mg of ammonia was dissolved in 3.0 l of aqueous solution. Then ammonium nitraternwas added to the same solution and the ph of the solution was found to be 9.8. if thernkb of ammonia is 1.8×10-5rn, calculate the concentration of ammonium ion in thisrnsolution.
Accepted Answer
Assuming the volume of the solution remains constant at 3.0 L after the addition of ammonium nitrate and neglecting any change in temperature, we can use the following equation to calculate the concentration of ammonium ion in the solution:
Kb = [NH4+][OH-] / [NH3]
where Kb is the base dissociation constant of ammonia, [NH4+] is the concentration of ammonium ion, [OH-] is the concentration of hydroxide ion, and [NH3] is the concentration of ammonia.
We can also use the following equation to relate the pH of the solution to the concentration of hydroxide ion:
pH = -log[OH-]
Substituting the given values into these equations, we get:
1.8 × 10^-5 = [NH4+][OH-] / 0.5 mg / 3.0 L
9.8 = -log[OH-]
Solving the second equation for [OH-], we get:
[OH-] = 10^-9.8 = 1.58 × 10^-10 M
Substituting this value into the first equation, we get:
1.8 × 10^-5 = [NH4+] × 1.58 × 10^-10 / 0.5 mg / 3.0 L
Solving for [NH4+], we get:
[NH4+] = 4.8 × 10^-6 M
Therefore, the concentration of ammonium ion in the solution is 4.8 × 10^-6 M.
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