Answer:
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
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:
[tex]0.02346 L \times \frac{0.238 mol}{L} = 5.58 \times 10^{-3} mol[/tex]
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).
[tex]5.58 \times 10^{-3} mol \times \frac{40.00g}{mol} = 0.223 g[/tex]
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.
[tex]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^{-}[/tex]
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:
[tex]5.58 \times 10^{-3} molNaOH \times \frac{1molH_2SO_4x}{2molNaOH} = 2.79\times 10^{-3} mol H_2SO_4[/tex]
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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