Chemistry, 08.09.2020 14:01 amandaestevez030

Consider a 0.238 M aqueous solution of sodium hydroxide, NaOH. a. How many grams of NaOH are dissolved in 23.46 mL?
b. How many individual hydroxide ions (OH) are found in 23.46 mL?
c. How many moles of sulfuric acid, H2SO4, are neutralized by 23.46 mL of 0.238 M NaOH(aq)?

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23.46 mL of 0.238 M contains 0.223 g of NaOH, 3.36 × 10²¹ OH⁻ ions and required 2.79 × 10⁻³ moles of sulfuric acid to be neutralized.

We have 23.46 mL of 0.238 M (mol/L) NaOH. The moles of NaOH are:

$0.02346 L \times \frac{0.238 mol}{L} = 5.58 \times 10^{-3} mol$

a. How many grams of NaOH are dissolved in 23.46 mL?

We can convert 5.58 × 10⁻³ moles of NaOH to grams using its molar mass (40.00 g/mol).

$5.58 \times 10^{-3} mol \times \frac{40.00g}{mol} = 0.223 g$

b. How many individual hydroxide ions (OH⁻) are found in 23.46 mL?

We can calculate the number of OH⁻ in 5.58 × 10⁻³ moles of NaOH using the following relations.

There is 1 mole of OH⁻ per mole of NaOH.There are 6.02 × 10²³ OH⁻ ions in 1 mole of OH⁻.

$5.58 \times 10^{-3} molNaOH \times \frac{1molOH^{-} }{1molNaOH} \times \frac{6.02 \times 10^{23}IonsOH^{-} }{1molOH^{-}} = 3.36 \times 10^{21}IonsOH^{-}$

c. How many moles of sulfuric acid, H₂SO₄, are neutralized by 23.46 mL of 0.238 M NaOH(aq)?

Let's consider the neutralization reaction between NaOH and H₂SO₄.

2 NaOH + H₂SO₄ ⇒ Na₂SO₄ + 2 H₂O

The molar ratio of NaOH to H₂SO₄ is 2:1. The moles of H₂SO₄ required to react with 5.58 × 10⁻³ moles of NaOH are:

$5.58 \times 10^{-3} molNaOH \times \frac{1molH_2SO_4x}{2molNaOH} = 2.79\times 10^{-3} mol H_2SO_4$

23.46 mL of 0.238 M contains 0.223 g of NaOH, 3.36 × 10²¹ OH⁻ ions and required 2.79 × 10⁻³ moles of sulfuric acid to be neutralized.

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Answer from: ltorline123

a. 0.223 grams of NaOH

b. 0.00558 moles of OH-

c. 0.00279 moles of H₂SO₄

Explanation:

0.238 M in a solution of NaOH means, that in 1 L of solution, we have 0.238 moles of NaOH. We can also say, that we have 0.238 mmoles in 1 mL of solution.

a. M = mmol / mL

mL . M = mmol → 5.58348 mmoles

We need to find the molar mass of NaOH → 40 mg /mmol

40 mg / mmol . 5.58348 mmol = 223.34 mg.

Let's convert to grams → 223.34 mg . 1g / 1000 mg = 0.223 g

b. NaOH → Na+ + OH-

Stoichiometry is 1:1, so in 23.46 mL we have 5.58 mmoles of base.

It is the same if we say, we have 5.58 mmoles of OH-.

We convert to moles → 5.58 mmol . 1 mol /1000mmol = 0.00558 mol

c. In the neutralization reaction, the moles of acid are the same as the moles of base

H₂SO₄ → 2H⁺ + SO4-2

NaOH → Na+ + OH-

In the stoychiometry we have 2 protons for 1 OH-, so the main formula will be:

2 mmol H+ = 1 mmol OH-

mmol H+ = 1 mmol OH- / 2

mmol H+ = (0.238 M . 23.46 mL) / 2

mmol H+ = 2.792 mmol

We convert to moles → 2.792 mmol . 1 mol /1000mmol = 0.00279 mol

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Consider a 0.238 M aqueous solution of sodium hydroxide, NaOH. a. How many grams of NaOH are dissol...

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