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2 years ago

7

Write an equilibrium expression for each chemical equation involving one or more solid or liquid reactants or products.

1 answer:

Alex_Xolod [135]2 years ago

5 0

Answer:

a.

$Keq=\frac{[HCO_3^-][OH^-]}{[CO_3^{2-}]}$

b.

$Keq=[O_2]^3$

c.

$Keq=\frac{[H_3O^+][F^-]}{[HF]}$

d.

$Keq=\frac{[NH_4^+][OH^-]}{[NH_3]}$

Explanation:

Hello there!

In this case, for the attached reactions, it turns out possible for us to write the equilibrium expressions by knowing any liquid or solid would be not-included in the equilibrium expression as shown below, with the general form products/reactants:

a.

$Keq=\frac{[HCO_3^-][OH^-]}{[CO_3^{2-}]}$

b.

$Keq=[O_2]^3$

c.

$Keq=\frac{[H_3O^+][F^-]}{[HF]}$

d.

$Keq=\frac{[NH_4^+][OH^-]}{[NH_3]}$

Regards!

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A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base

tamaranim1 [39]

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ = 0.00375 mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

$HN_3 + OH- ---> N^-_{3} + H_2O$

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thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L

Concentration of $HN_3$ = $\dfrac{0.001755}{0.0383}$ = 0.0458 M

Concentration of $N^{-}_3$ = $\dfrac{ 0.001995 }{0.0383}$ = 0.0521 M

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Ka = $1.9 x 10^{-5}$

Thus; it's pKa = 4.72

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Explanation:

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