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denis23 [38]
2 years ago
9

A student is asked to standardize a solution of barium hydroxide. He weighs out 0.978 g potassium hydrogen phthalate (KHC8H4O4,

treat this as a monoprotic acid). It requires 35.8 mL of barium hydroxide to reach the endpoint. A. What is the molarity of the barium hydroxide solution? M This barium hydroxide solution is then used to titrate an unknown solution of hydrochloric acid. B. If 17.1 mL of the barium hydroxide solution is required to neutralize 18.6 mL of hydrochloric acid, what is the molarity of the hydrochloric acid solution? M
Chemistry
1 answer:
Sav [38]2 years ago
3 0

Answer:

(A) 0.129 M

(B) 0.237 M

Explanation:

(A) The reaction between potassium hydrogen phthalate and barium hydroxide is:

  • 2HA + Ba(OH)₂ → BaA₂ + 2H₂O

Where A⁻ is the respective anion of the monoprotic acid (KC₈H₄O₄⁻).

We <u>convert mass of phthalate to moles</u>, using its molar mass:

  • 0.978 g ÷ 156 g/mol = 9.27x10⁻³ mol = 9.27 mmol

Now we <u>convert mmol of HA to mmol of Ba(OH)₂</u>:

  • 9.27 mmol HA * \frac{1mmolBa(OH)_{2}}{2mmolHA} = 6.64 mmol Ba(OH)₂

Finally we calculate the molarity of the Ba(OH)₂ solution:

  • 6.64 mmol / 35.8 mL = 0.129 M

(B) The reaction between Ba(OH)₂ and HCl is:

  • 2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O

So<u> the moles of HCl that reacted </u>are:

  • 17.1 mL * 0.129 M * \frac{2mmolHCl}{1mmolBa(OH)_2} = 4.41 mmol HCl

And the <u>molarity of the HCl solution is</u>:

  • 4.41 mmol / 18.6 mL = 0.237 M

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Answer:

The statements which are true among these are: (a),(b) and (c) because,

(a) The simplest organic compounds which contains only carbon and hydrogen atoms are called hydrocarbons.

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Rhodium crystallizes in a face-centered cubic unit cell. The radius of a rhodium atom is 135 pm. Determine the density of rhodiu
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Density of unit cell ( rhodium) = 12.279 g/cm³

Explanation:

Given that:

The radius (r) of a rhodium atom = 135 pm

The atomic mass of rhodium = 102.90 amu

For a face-centered cubic unit cell,

r = \dfrac{a}{2\sqrt{2}}

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a = 2 \times 1.414 \times 135 \ pm

a = 381.8 pm

to cm, we get:

a = 381.8 × 10⁻¹⁰ cm

However, recall that:

density \ of \ unit \ cell = \dfrac{mass \ of \ unit \ cell}{volume \ of \unit \ cell}

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mass of unit cell = mass of atom × numbers of atoms per unit cell

Also;

mass\  of\ atom =\dfrac{ atomic \ mass}{Avogadro  \  number}

mass\  of\ atom =\dfrac{ 102.9}{6.023 \times 10^{23}}

Recall also that number of atoms in a unit cell for a  face-centered cubic = 4

So;

mass \ of \ unit \ cell= \dfrac{102.90}{6.023 \times 10^{23}}\times 4

mass of unit cell = 6.83380375 × 10⁻²² g

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