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vazorg [7]
3 years ago
11

Write complete ionic equation to show the reaction of aqueous lead(II) nitrate with aqueous potassium sulfate to form solid lead

(II) sulfate and aqueous potassium nitrate.
Chemistry
2 answers:
leva [86]3 years ago
8 0
The correct answer to this question is this one:

<span>Pb(N<span>O3</span><span>)2</span>(aq)+<span>K2</span>S<span>O4</span>(aq)→PbS<span>O4</span>(s)+2KN<span>O3</span>(aq)
</span>Everything except the lead sulfate is soluble, so they all dissociate.

<span>P<span>b<span>2+</span></span>+2N<span>O−3</span>+2<span>K+</span>+S<span>O<span>2−</span>4</span>→PbS<span>O4</span>(s)+2<span>K+</span>+2N<span>O−3
</span></span>Cancel out stuff that appears on both sides, and your net ionic equation is

<span>P<span>b<span>2+</span></span>(aq)+S<span>O<span>2−</span>4</span>(aq)→PbS<span>O4</span>(s)</span>
Semenov [28]3 years ago
8 0

Answer:

Pb²⁺ (aq) + 2NO₃⁻ (aq) + 2K⁺ (aq) + SO₄²⁻ (aq) → PbSO₄(s) + 2K⁺ + 2 NO₃⁻ (aq)

Explanation:

1) Start understanding the nature of the reactants, the kind of reaction that occurs, and the nature of the products:

i) word equation (given):

aqueous lead(II) nitrate + aqueous potassium sulfate → solid lead(II) sulfate + aqueous potassium nitrate

ii) molecular equation, including the phases:

Pb(NO₃)₂ (aq) + K₂SO₄ (aq) → PbSO₄(s) + 2KNO₃ (aq)

iii) It is a combination of two salts to form two new salts, in a double replacement reaction.

2) Separate the ionic compounds in solution into its ions

i) Reactant side:

Pb(NO₃)₂ (aq) = Pb²⁺ (aq) + 2NO₃⁻ (aq)

K₂SO₄ (aq) = 2K⁺ (aq) + SO₄²⁻ (aq)

ii) Product side

PbSO₄(s) remains unchanged since it did not ionize

2KNO₃(aq) = 2K⁺ + 2 NO₃⁻ (aq)

3) Write both sides, i.e. the whole reaction in the ionic form:

Pb²⁺ (aq) + 2NO₃⁻ (aq) + 2K⁺ (aq) + SO₄²⁻ (aq) → PbSO₄(s) + 2K⁺ + 2 NO₃⁻ (aq)

That is the total ionic equation.

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Read 2 more answers
A student prepares a aqueous solution of butanoic acid . Calculate the fraction of butanoic acid that is in the dissociated form
ella [17]

Answer:

15.4%

Explanation:

If Ka = 0.54 mM = 1.51x10⁻⁵

Then;

C₄H₈O₂               -------->            C₄H₇O₂⁻          +           H⁺

I                    0.54x10⁻³                             0                                0

E                   0.54x10⁻³(1-x)                      0.54x10⁻³x                0.54x10⁻³x

Recall that x is the percentage degree of dissociation

From the ICE table;

Ka = [C₄H₇O₂⁻] [ H⁺]/[C₄H₈O₂]

1.51x10⁻⁵=(0.54x10⁻³x) (0.54x10⁻³x)/ 0.54x10⁻³(1-x)  

1.51x10⁻⁵ = 0.54x10⁻³x^2/1-x

1.51x10⁻⁵(1-x) = 0.54x10⁻³x^2

1.51x10⁻⁵ - 1.51x10⁻⁵x = 0.54x10⁻³x^2

Hence;

0.54x10⁻³x^2 + 1.51x10⁻⁵x - 1.51x10⁻⁵=0

x^2 + 0.028x - 0.028 = 0

Solving the quadratic equation here;

x = 0.154 or −0.182

Ignoring the negative result, x = 0.154

Hence, fraction of butanoic acid that is in the dissociated form in this solution = 15.4%

3 0
3 years ago
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