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blagie [28]
3 years ago
15

An aqueous solution of ________ will produce a neutral solution. An aqueous solution of ________ will produce a neutral solution

. NaNO2 Li2CO3 NaF NaClO4 NH4ClO4
Chemistry
1 answer:
IRINA_888 [86]3 years ago
8 0

Answer:

An aqueous solution of  NaF will produce a neutral solution.

Explanation:

In order to classify the solutions, we need to dissociate all the salts and determine the conjugate pairs of base and acids.

NaNO₂ → Na⁺  + NO₂⁻

Na⁺ comes from the NaOH, a strong base, so the Na⁺ is the conjugate weak acid from the strong base. It will not react. This is a basic salt.

The NO₂⁻  is a conjugate strong base, that comes from the weak acid, nitrous acid. This ion can make hydrolisis.

NO₂⁻  +  H₂O  ⇄    HNO₂   +  OH⁻          Kb

We give OH⁻ to medium, so solution is basic.

Li₂CO₃  →  2Li⁺  +  CO₃⁻²

Li⁺ comes from the LiOH, a strong base, so the cathion is the conjugate weak acid from the strong base. It will not react

The CO₃⁻²  is a conjugate strong base, that comes from the weak acid, carbonic acid. This ion can make hydrolisis.

CO₃⁻²  +  H₂O  ⇄    HCO₃⁻  +  OH⁻      Kb

NaClO₄ →  Na⁺  +  ClO₄⁻

Na⁺ comes from the NaOH, a strong base, so the cathion is the conjugate weak acid from the strong base. It will not react

The ClO₄⁻  is a conjugate strong base, that comes from the weak acid, perchloric acid. This ion can make hydrolisis. This is a basic salt.

ClO₄⁻  +  H₂O  ⇄    HClO₄⁻  +  OH⁻      Kb

NH₄ClO₄  →  NH₄⁺  +  ClO₄⁻

Both ions can make hydrolisis. Ammonium comes from a weak base, so it is the strong conjugate base, and  ClO₄⁻  is a conjugate strong base, that comes from the weak acid, perchloric acid

NH₄⁺  +  H₂O  ⇄  NH₃  +  H₃O⁺         Ka

ClO₄⁻  +  H₂O  ⇄    HClO₄⁻  +  OH⁻      Kb

This sort of salt are generally acid.

NaF  →  Na⁺  + F⁻

As we have seen, Na⁺ comes from the NaOH (a strong base) so it will not react.

This equilibrium can not occur: Na⁺  + H₂O ←  NaOH   +  H⁺

F⁻  comes from a strong acid, HF so it will not also react. This reaction can not occur too: F⁻  + H₂O ←  HF   +  OH⁻

We do not have reaction in this neutral salt, so the pH is neutral.

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A) Calculate the osmotic pressure difference between seawater and fresh water. For simplicity, assume thatall the dissolved salt
never [62]

Answer:

a)  Δπ = 1.264 atm

b) W = 128 joules

c)  ΔH >> W  ( a factor greater than 17,000 )

Explanation:

a) The osmotic pressure, π , is determined by :

π = nRT/V, where n= moles of solute

                          R= 0.0821 Latm/kmol

                          T = 300 K

calling π(sw) osmotic pressure for  for sea water and π (fw) for fresh water,

salinity of sea water = 3.5 g / 1L water   (assuming only NaCl for the salts)

salinity of fresh water = 0.5 parts per thousand (range: 0- 0.5 ppt)

πsw = (3.5 g/58.44 g/mol) (0.0821 Latm/Kmol) (300 K ) /1 L = 1.475 atm

πfw = (0.5 g/58.44 g/mol) (0.0821 Latm/Kmol) (300 K ) /1 L = 0.211 atm

d water = 1 g/cm³

Δ π = (1.475 - 0.211) = 1.264 atm

b) W = Δπ V = 1.426 atm x 1L = 1.43 L-atm

1 L-atm = 101.33 j

W =  101.33 j/ Latm x  1.43 Latm = 128 joules

c) ΔH = Q₁ + nΔH vap, where

            Q₁  = heat required to bring the solution from 300 K to boiling, 373 K

            ΔH vap = heat of vaporization

Q = mCΔT = 1000 g x 4.186 j x 73 K = 305.6 j = 0.3056 kj

ΔH vap = (1000 g/ 18 g/mol ) 40.7 kj/mol = 2,261 kj

ΔH =  0.3056 kj + 2,261 kj = 2,261.3 kj

Note = Q << ΔH vap and we could have neglected it.

This result shows why nobody talks about evaporation of sea water to produce fresh water ΔH >> W

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<h2>Answer:</h2>

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<h3>Explanation:</h3>

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