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Inessa05 [86]
3 years ago
7

what is the molarity of 20.0 ml of a KCl solution that reacts completely with 30.0 ml of a 0.400 M Pb(NO3)2 solution?

Chemistry
1 answer:
Alex17521 [72]3 years ago
7 0

Answer:

[KCl] = 1.2 M

Explanation:

We need to complete the reaction:

2KCl(aq) + Pb(NO₃)₂(aq)  → 2KNO₃(aq) + PbCl₂(s)↓

By stoichiomety we know that 1 mol of chloride needs 1 mol of nitrate to react:

Let's find out the moles of nitrate, we have:

Molarity = mol/volume(L)

We convert the volume → 30 mL . 1L/1000mL = 0.030L

Molarity . volume(L) = moles → 0.400 M . 0.030L = 0.012 moles

Therefore, we can make a rule of three.

1 mol of nitrate reacts with 2 moles of chloride

Then, 0.012 moles of nitrate must react with (0.012 . 2) / 1 = 0.024 moles of KCl

We convert the volume from mL to L → 20 mL . 1L /1000mL = 0.020L

Molarity = mol /volume(L) → 0.024 mol /0.020L = 1.2 M

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a)- The chemical equation for the corresponden equilibrium of Ka1 is:

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c)- According to the previous demonstation, at equilibrium ΔG= 0.

d)- In a non-equilibrium condition, we have Q which is calculated with the concentrations of products and reactions in a non equilibrium state:

ΔG= ΔGº + RT ln Q

Q= ((H⁺) (NO₂⁻))/(HNO₂)

Q= ( (5.9 10⁻² M) x (6.7 10⁻⁴ M) ) / (0.21 M)

Q= 1.88 10⁻⁴

We know that   ΔGº= 19092.8 J/mol, so:

ΔG= ΔGº + RT ln Q

ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)

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