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Angelina_Jolie [31]
3 years ago
8

What is the pH of 0.26 M ammonium ion? NH4+(aq) + H2O(1) NH3(aq) + H30* (aq) a. 4.33 b.9.25 c. 3.87 d. 4.92 e. 4.75

Chemistry
1 answer:
choli [55]3 years ago
6 0

Answer:

b) pH = 9.25

Explanation:

  • NH4+(aq)  +  H2O(l)  ↔  NH3(aq)  +  H3O+(aq)
  • NH3 + H2O ↔ NH4+  +  OH-
  • 2 H2O ↔ H3O+  +  OH-

⇒ Kb = [ NH4+ ] * [ OH- ] / [ NH3 ] = 1.86 E-5......from literature

mass balance NH4+:

⇒ M NH4+ = [ NH4+ ] - [ OH- ]

∴ [ NH3 ] ≅ M NH4+ = 0.26 M

⇒ Kb = (( 0.26 + [ OH- ] )) * [ OH- ] / 0.26 = 1.86 E-5

⇒ 0.26 [ OH-] + [ OH- ]² = 4.836 E-6

⇒ [ OH- ]² + 0.26 [ OH- ] - 4.836 E-6 = 0

⇒ [ OH- ] = 1.859 E-5 M

⇒ pOH = - Log ( 1.859 E-5 )

⇒ pOH = 4.7305

⇒ pH = 14 - pOH = 9.269

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andrey2020 [161]

Answer:

V_2=8L

Explanation:

Hello there!

In this case, considering the Avogadro's gas law, which treats the volume and moles in a directly proportional way via:

\frac{V_1}{n_1}=\frac{V_2}{n_2}

Which can be solved for the final volume, V2, as shown below:

V_2=\frac{V_1n_2}{n_1}

Thus, by plugging in the given data, we obtain:

V_2=\frac{4L*(0.250mol+0.250mol)}{0.250mol}\\\\V_2=8L

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4 0
2 years ago
PLEASE HELP QUICKLY!! How many moles of NH4OH are in 0.125L of a 1.4 M solution of NH4OH?
frez [133]

Answer:

0.175mol

Explanation:

Molarity of a solution can be calculated thus:

Molarity (M) = number of moles (n) / volume (V)

According to the information provided in this question, volume of NH4OH = 0.125L, molarity = 1.4M

Using; Molarity = n/V

1.4 = n/0.125

n = 0.175mol

5 0
3 years ago
Consider the reaction: 2HI(g) ⇄ H2(g) + I2(g). It is found that, when equilibrium is reached at a certain temperature, HI is 35.
Sliva [168]

Answer:

Kc = 0.075

Explanation:

The dissociation (α) is the initial quantity that ionized divided by the total dissolved. So, let's calling x the ionized quantity, and M the initial one:

α = x/M

x = M*α

x = 0.354M

For the stoichiometry of the reaction (2:1:1), the concentration of H₂ and I₂ must be half of the acid. So the equilibrium table must be:

2HI(g) ⇄   H₂(g) +    I₂(g)

M               0             0               <em> Initial</em>

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The equilibrium constant Kc is the multiplication of the products' concentrations (elevated by their coefficients) divided by the multiplication of the reactants' concentrations (elevated by their coefficients):

Kc = \frac{[H2]*[I2]}{[HI]^2}

Kc = \frac{0.177M*0.177M}{(0.646M)^2}

Kc = \frac{0.03133M^2}{0.41732M^2}

Kc = 0.075

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