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svp [43]
2 years ago
10

Potassium hydrogen phthalate is often used to standardize solutions of strong bases. Highly pure samples of this solid weak acid

are readily available. It is stable and easy to use in the laboratory. Calculate the molarity of a sodium hydroxide solution if 23.73 mL are required to titrate 0.5816 g of potassium hydrogen phthalate (C8H5O4K).
Chemistry
1 answer:
Mazyrski [523]2 years ago
3 0

Answer:

0.12 M

Explanation:

Molarity of sodium hydroxide = number of moles of sodium hydroxide / volume in liters

equation of reaction between NaOH and Potassium hydrogen phthalate

NaOH(aq) + KHC₈H₄O₄(aq)  KNaC₈H₄O₄(aq) + H₂O(aq)

molar mass of KHC₈H₄O₄ = 204.22g/mol

amount required to titrate with NaOH = 0.5816 g

mole of  KHC₈H₄O₄ = 0.5816 g / 204.22g/mol = 0.00285 mole

considering the equation, 1 mole of NaOH required 1 mole of KHC₈H₄O₄

0.00285 mole  KHC₈H₄O₄ will require 0.00285 mole of NaOH

Molarity = 0.00285 mole  / ( 23.73 ml / 1000ml) L = 0.12 M

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<u>Answer:</u> The average atomic mass of X is 28.09 amu

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i .....(1)

  • <u>For isotope 1:</u>

Mass of isotope 1 = 27.979 amu

Percentage abundance of isotope 1 = 92.21 %

Fractional abundance of isotope 1 = 0.9212

  • <u>For isotope 2:</u>

Mass of isotope 2 = 28.976 amu

Percentage abundance of isotope 2 = 4.70 %

Fractional abundance of isotope 2 = 0.0470

  • <u>For isotope 3:</u>

Mass of isotope 3 = 29.974 amu

Percentage abundance of isotope 3 = 3.09 %

Fractional abundance of isotope 3 = 0.0309

Putting values in equation 1, we get:

\text{Average atomic mass of X}=[(27.979\times 0.9212)+(28.976\times 0.0470)+(29.974\times 0.0309)]

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4 0
3 years ago
What is the pH of 0.10 M NaF(aq). The Ka of HF is 6.8 x 10-4
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<u>Given information:</u>

Concentration of NaF = 0.10 M

Ka of HF = 6.8*10⁻⁴

<u>To determine:</u>

pH of 0.1 M NaF

<u>Explanation:</u>

NaF (aq) ↔ Na+ (aq) + F-(aq)

[Na+] = [F-] = 0.10 M

F- will then react with water in the solution as follows:

F- + H2O ↔ HF + OH-

Kb = [OH-][HF]/[F-]

Kw/Ka = [OH-][HF]/[F-]

At equilibrium: [OH-]=[HF] = x and [F-] = 0.1 - x

10⁻¹⁴/6.8*10⁻⁴ = x²/0.1-x

x = [OH-] = 1.21*10⁻⁶ M

pOH = -log[OH-] = -log[1.21*10⁻⁶] = 5.92

pH = 14 - pOH = 14-5.92 = 8.08

Ans: (b)

pH of 0.10 M NaF is 8.08

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