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In a titration of 35.00 ml of 0.737 m h2so4, __________ ml of a 0.827 m koh solution is required for neutralization.

Respuesta :

the reaction between H₂SO₄ and KOH is as follows;
2KOH  + H₂SO₄ --> K₂SO₄  + 2H₂O
stoichiometry of KOH to H₂SO₄ is 2:1
the number of H₂SO₄ moles that have reacted = 0.737 M /1000mL x 35 mL 
                                                                          = 0.0258 mol
According to stoichiometry, number of KOH moles that have reacted at point of neutralization = number of H₂SO₄ moles x 2
Therefore number of KOH moles = 0.0516 mol
The molarity of KOH = 0.827 mol/L 
The volume of 0.827 mol - 1 L 
Therefore volume for 0.0516  mol  - 1/0.827 x 0.0516 = 0.0624 L
volume of KOH needed = 62.4 mL 

In a titration of 35.00 ml of 0.737 M sulfuric acid, 62.38 ml of a 0.827 M KOH solution is required for neutralization.

In the neutralization, reaction of [tex]\rm H_2SO_4[/tex] and KOH, the reaction will be:

[tex]\rm H_2SO_4\;+2\;KOH\;\rightarrow\;K_2SO_4\;+\;2\;H_2O[/tex]

For 1 mole of [tex]\rm H_2SO_4[/tex] 2 moles of KOH is required.

Moles of [tex]\rm H_2SO_4[/tex] IN 0.737 M solution :

Molarity = [tex]\rm moles\;\times\;\dfrac{1000}{Volume\;(ml)}[/tex]

moles = [tex]\rm molarity\;\times\;\dfrac{volume\;(ml)}{1000}[/tex]

Moles of  [tex]\rm H_2SO_4[/tex] = 0.737 [tex]\rm \times\;\dfrac{35}{1000}[/tex]

Moles of  [tex]\rm H_2SO_4[/tex] = 0.0257 moles.

Moles of KOH required = 2 [tex]\times[/tex] moles of  [tex]\rm H_2SO_4[/tex]

Moles of KOH required = 2 [tex]\times[/tex] 0.257

Moles of KOH required = 0.0515 moles

Volume of 0.827 M KOH will be:

Volume (ml) = [tex]\rm \dfrac{moles}{molarity}\;\times\;1000[/tex]

Volume (ml) = [tex]\rm \dfrac{0.0515}{0.827}\;\times\;1000[/tex]

Volume of KOH = 62.38 ml

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https://brainly.com/question/8732513?referrer=searchResults

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