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quester [9]
3 years ago
11

Is sodium benzoate soluble in NaOH, NaHCO3

Chemistry
2 answers:
krek1111 [17]3 years ago
6 0
It is water soluble so is also soluble in aqueous solutions of NaOH or NaHCO3.
storchak [24]3 years ago
6 0

Answer:

It's slightly soluble in an aqueous solution of NaHCO_{3}, and almost insoluble in an aqueous solution of NaOH.

Explanation:

Sodium benzoate comes from benzoic acid, which is a weak acid. It means that in an aqueous solution benzoic acid does not ionize easily to form the ions H_{3}O^{+} and C_{7}H_{5}O_{2} ^{-}

It also implies, according to the Le Châtelier's principle, that the ion C_{7}H_{5}O_{2}^{-} tends to form the acid C_{7}H_{6} O_{2} more easily. It can be seen in the following equation:

C_{7}H_{6} O_{2}  ⇔ C_{7}H_{5}O_{2}^{-}  + H_{3}O^{+}

In an aqueous solution, the equilibrium shifts to the left, thus letting water dissolve sodium benzoate.  But why? Because water in that case would produce enough H_{3}O^{+} ions to facilitate the disolution of sodium benzoate. It's shown by its solubility in water at 15°C (62.78g/100mL, according to Wikipedia).

In contrast, the presence of NaOH or NaHCO_{3}, both chemical species producing the OH^{-} ions in aqueous solution, would make the equilibrium shift to the right because it would be a higher need of H_{3}O^{+} ions to offset the presence of OH^{-}.

However, the effect of NaOH is not the same due to NaHCO_{3}, because the first is a strong base and the other is a weak one. Thereby it is reasonable to think that solubility of sodium benzoate is greater in water than in NaHCO_{3} and NaOH.

Solubility in water > solubility in  NaHCO_{3}> solubility in NaOH.

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3 0
1 year ago
How many milliliters of 0.200 m fecl3 are needed to react with an excess of na2s to produce 1.38 g of fe2s3 if the percent yield
ahrayia [7]
<span>Answer: 100 ml
</span>

<span>Explanation:


1) Convert 1.38 g of Fe₂S₃ into number of moles, n


</span>i) Formula: n = mass in grass / molar mass
<span>
ii) molar mass of </span><span>Fe₂S₃ =2 x 55.8 g/mol + 3 x 32.1 g/mol = 207.9 g/mol
</span>

iii) n = 1.38 g / 207.9 g/mol = 0.00664 moles of <span>Fe₂S₃
</span>

<span>2) Use the percent yield to calculate the theoretical amount:
</span>

<span>65% = 0.65 = actual yield/ theoretical yield =>


</span>theoretical yield = actual yield / 0.65 = 0.00664 moles / 0.65 = 0.010 mol <span>Fe₂S₃</span><span>

3) Chemical equation:
</span>

<span> 3 Na₂S(aq) + 2 FeCl₃(aq) → Fe₂S₃(s) + 6 NaCl(aq)


4) Stoichiometrical mole ratios:
</span>

<span>3 mol Na₂S : 2 mol FeCl₃ : 1 mol Fe₂S₃ : 6 mol NaCl


5) Proportionality:


</span>2moles FeCl₃ / 1 mol Fe₂S₃ = x / 0.010 mol Fe₂S₃
<span>
=> x = 0.020 mol FeCl₃


6) convert 0.020 mol to volume
</span>

<span>i) Molarity formula: M = n / V
</span>

<span>ii) V = n / M = 0.020 mol / 0.2 M = 0.1 liter = 100 ml
</span>

3 0
3 years ago
Read 2 more answers
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