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Leya [2.2K]
3 years ago
11

what is the pH of a solution that results when 0.010mol HNO3 is added to 500.ml of a solution that is 0.10M in aqueous ammonia a

nd 0.55 M in ammonium nitrate. assume no volume change. (The Kb for NH3 =1.8 * 10-5 )
Chemistry
1 answer:
WARRIOR [948]3 years ago
8 0

Answer : The  

pH of a solution is, 8.56

Explanation : Given,

K_b=1.8\times 10^{-5}

Concentration of ammonia (base) = 0.10 M

Concentration of ammonium nitrate (salt) = 0.55 M

First we have to calculate the value of pK_b.

The expression used for the calculation of pK_b is,

pK_b=-\log (K_b)

Now put the value of K_b in this expression, we get:

pK_b=-\log (1.8\times 10^{-5})

pK_b=5-\log (1.8)

pK_b=4.7

Now we have to calculate the pOH of buffer.

Using Henderson Hesselbach equation :

pOH=pK_b+\log \frac{[Salt]}{[Base]}

Now put all the given values in this expression, we get:

pOH=4.7+\log (\frac{0.55}{0.10})

pOH=5.44

The pOH of buffer is 5.44

Now we have to calculate the pH of a solution.

pH+pOH=14\\\\pH+5.44=14\\\\pH=14-5.44\\\\pH=8.56

Thus, the pH of a solution is, 8.56

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2Pb(s) + O2(aq) + 4H+(aq) → 2H2O(l) + 2Pb2+(aq)
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Answer:

The answer to your question is 0.269 g of Pb

Explanation:

Data

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mass = 0.269 g

atomic mass Pb = 207.2 g

Chemical reaction

                        2Pb(s) + O₂(aq) + 4H⁺(aq) → 2H₂O(l) + 2Pb₂⁺(aq)

Process

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                       207.2 g of Pb ----------------- 1 mol

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