The molarity of the sodium hydroxide is calculated by first calculating the number of moles of sulfuric acid in the 22.30 ml used.
This is done as follows since 0.253M is contained in 1000ml solution.
(0.253×22.3)/1000=0.00564 moles
The equation for the reaction is:
H₂SO₄₍ₐq)+2NaOH₍ₐq)⇒ Na₂SO₄₍aq) +2 H₂O₍l₎
Therefore the reacting ratios between sodium hydroxide and sulfuric acid is 2:1
therefore the number of moles that reacted with the sulfuric acid is calculated as follows:
(0.00564 moles ₓ 2)/1=0.01128moles
0.1128moles is in 25ml therefore, 1000ml has:
(1000ₓ0.01128moles)/25= 0.4512M NaOH
22.30 mL of 0.253 M H₂SO₄ titrates 25.0 mL of 0.448 M NaOH.
Let's consider the neutralization reaction between sulfuric acid and sodium hydroxide.
2 NaOH + H₂SO₄ ⇒ Na₂SO₄ + 2 H₂O
22.30 mL of 0.253 M H₂SO₄ react. The reacting moles of H₂SO₄ are:
[tex]0.02230 L \times \frac{0.253mol}{L} = 5.64 \times 10^{-3} mol[/tex]
The molar ratio of NaOH to H₂SO₄ is 2:1. The reacting moles of NaOH are:
[tex]5.64 \times 10^{-3} mol H_2SO_4 \times \frac{2molNaOH}{1 molH_2SO_4} = 0.0112 molNaOH[/tex]
0.0112 moles of NaOH are in 25.0 mL of solution. The molarity of NaOH is:
[tex]M = \frac{0.0112 mol}{0.0250L} = 0.448 M[/tex]
22.30 mL of 0.253 M H₂SO₄ titrates 25.0 mL of 0.448 M NaOH.
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