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Andreas93 [3]
2 years ago
13

How much 1.5 m sodium hydroxide is necessary to exactly neutralize 20.0 ml of 2.5 m phosphoric acid?

Chemistry
1 answer:
andrew-mc [135]2 years ago
8 0
Answer is: 100 mL of <span>sodium hydroxide.
Chemical reaction: 3NaOH + H</span>₃PO₄ → Na₃PO₄ + 3H₂O.
V(H₃PO₄) = 20.0 mL ÷ 1000 mL/L = 0.02 L.
n(H₃PO₄) = V(H₃PO₄) · c(H₃PO₄).
n(H₃PO₄) = 0.02 L · 2.5 mol/L.
n(H₃PO₄) = 0.05 mol.
From chemical reaction: n(H₃PO₄) : n(NaOH) = 1 : 3.
n(NaOH) = 0.15 mol.
V(NaOH) = n(NaOH) ÷ c(NaOH).
V(NaOH) = 0.15 mol ÷ 1.5 mol/L.
V(NaOH) = 0.1 L · 1000 mL/L = 100 mL.
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Answer : The molarity of the chloride ion in the water is, 5.75 M

Explanation :

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First we have to calculate the volume of solution.

\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}

\text{Volume of solution}=\frac{100g}{1.23g/mL}=81.3mL

Now we have to calculate the molarity of chloride ion.

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Formula used :

\text{Molarity}=\frac{\text{Mass of chloride ion}\times 1000}{\text{Molar mass of chloride ion}\times \text{Volume of solution (in mL)}}

Now put all the given values in this formula, we get:

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Thus, the molarity of the chloride ion in the water is, 5.75 M

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<em>Answer: 5.85 oz. </em>
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