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elena-14-01-66 [18.8K]
3 years ago
13

If 25.0 mL of 0.451 M NaOH solution is titrated with 0.253 M H2SO4, the flask at the endpoint will contain (besides the indicato

r phenolphthalein) as the principal components:
Chemistry
1 answer:
tigry1 [53]3 years ago
4 0

Answer:

H₂O, Na⁺, and SO₄²⁻

Explanation:

NaOH and H₂SO₄ are strong electrolytes. They will ionize completely .

Molecular equation:          2NaOH + H₂SO₄ ⟶ Na₂SO₄ + 2H₂O

Ionic equation: 2Na⁺ + 2OH⁻ + 2H⁺ + SO₄²⁻ ⟶  2Na⁺ + SO₄²⁻+ 2H₂O

  • The concentration of water is 55.5 mol·L⁻¹.
  • The concentration of Na⁺ is twice that of SO₄²⁻ .
  • Hydrolysis of SO₄²⁻ is negligible.

The principal components in the flask at the endpoint are H₂O, Na⁺, and SO₄²⁻.

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A solution was prepared by dissolving 195.0 g of KCl in 215 g of water. Calculate the mole fraction of KCl. (The formula weight
defon

Answer:

Approximately 0.180.

Explanation:

The mole fraction of a compound in a solution is:

\displaystyle \frac{\text{Number of moles of compound in question}}{\text{Number of moles of all particles in the solution}}.

In this question, the mole fraction of \rm KCl in this solution would be:

\displaystyle X_\mathrm{KCl} = \frac{n(\mathrm{KCl})}{n(\text{All particles in this soluton})}.

This solution consist of only \rm KCl and water (i.e., \rm H_2O.) Hence:

\begin{aligned} X_\mathrm{KCl} &= \frac{n(\mathrm{KCl})}{n(\text{All particles in this soluton})}\\ &= \frac{n(\mathrm{KCl})}{n(\mathrm{KCl}) + n(\mathrm{H_2O})}\end{aligned}.

From the question:

  • Mass of \rm KCl: m(\mathrm{KCl}) = 195.0\; \rm g.
  • Molar mass of \rm KCl: M(\mathrm{KCl}) = 74.6\; \rm g \cdot mol^{-1}.
  • Mass of \rm H_2O: m(\mathrm{H_2O}) = 215\; \rm g.
  • Molar mass of \rm H_2O: M(\mathrm{H_2O}) = 18.0\; \rm g\cdot mol^{-1}.

Apply the formula \displaystyle n = \frac{m}{M} to find the number of moles of \rm KCl and \rm H_2O in this solution.

\begin{aligned}n(\mathrm{KCl}) &= \frac{m(\mathrm{KCl})}{M(\mathrm{KCl})} \\ &= \frac{195.0\; \rm g}{74.6\; \rm g \cdot mol^{-1}} \approx 2.61\; \em \rm mol\end{aligned}.

\begin{aligned}n(\mathrm{H_2O}) &= \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} \\ &= \frac{215\; \rm g}{18.0\; \rm g \cdot mol^{-1}} \approx 11.9\; \em \rm mol\end{aligned}.

The molar fraction of \rm KCl in this solution would be:

\begin{aligned} X_\mathrm{KCl} &= \frac{n(\mathrm{KCl})}{n(\text{All particles in this soluton})}\\ &= \frac{n(\mathrm{KCl})}{n(\mathrm{KCl}) + n(\mathrm{H_2O})} \\ &\approx \frac{2.61 \; \rm mol}{2.61\; \rm mol + 11.9\; \rm mol} \approx 0.180\end{aligned}.

(Rounded to three significant figures.)

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