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Aneli [31]
3 years ago
6

Write the balanced chemical equation between H2SO4 and KOH in aqueous solution. This is called a neutralization reaction and wil

l produce whater and potassium sulfate. Phases are optional.
0.650 L of 0.430 M H2SO4 is mixed with 0.600 L of 0.240 M KOH. What concentration of sulfuric acid remains after neutralization?
Chemistry
1 answer:
emmainna [20.7K]3 years ago
7 0

Answer:

0.166M

Explanation:

In a neutralization, the acid, H₂SO₄, reacts with a base, KOH, to produce a salt, K₂SO₄ and water. The reaction is:

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

To solve this problem, we need to determine moles of H2SO4 and moles of KOH that reacts to find the moles of sulfuric acid that remains after the reaction:

<em>Moles H2SO4:</em>

0.650L * (0.430mol /L) = 0.2795moles H2SO4

<em>Moles KOH:</em>

0.600L * (0.240mol / L) = 0.144 moles KOH

Moles of sulfuric acid that reacts with 0.144 moles of KOH are:

0.144 moles KOH * (1mol H2SO4 / 2 mol KOH) = 0.072 moles of H2SO4 react.

And remain:

0.2795moles H2SO4 - 0.072moles H2SO4 = 0.2075 moles of H2SO4 reamains.

In 0.650L + 0.600L = 1.25L:

Molar concentration of sulfuric acid:

0.2075 moles of H2SO4 / 1.25L =

<h3>0.166M</h3>
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A chemistry graduate student is given of a chlorous acid solution. Chlorous acid is a weak acid with . What mass of should the s
DerKrebs [107]

Answer:

11.31g NaClO₂

Explanation:

<em> Is given 250mL of a 1.60M chlorous acid HClO2 solution. Ka is 1.110x10⁻². What mass of NaClO₂ should the student dissolve in the HClO2 solution to turn it into a buffer with pH =1.45? </em>

It is possible to answer this question using Henderson-Hasselbalch equation:

pH = pKa + log₁₀ [A⁻] / [HA]

<em>Where pKa is -log Ka = 1.9547; [A⁻] is the concentration of the conjugate base (NaClO₂), [HA] the concentration of the weak acid</em>

You can change the concentration of the substance if you write the moles of the substances:

[Moles HClO₂] = 250mL = 0.25L×(1.60mol /L) = <em>0.40 moles HClO₂</em>

Replacing in H-H expression, as the pH you want is 1.45:

1.45 = 1.9547 + log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]

-0.5047 = log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]

<em>0.3128 = </em>[Moles NaClO₂] / [0.40 moles HClO₂]

0.1251 = Moles NaClO₂

As molar mass of NaClO₂ is 90.44g/mol, mass of 0.1251 moles of NaClO₂ is:

0.1251 moles NaClO₂ ₓ (90.44g / mol) =

<h3>11.31g NaClO₂</h3>
5 0
3 years ago
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