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3 years ago

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!

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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

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1.4 = n/0.125

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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

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x = 0.354M

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M 0 0 Initial

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