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beks73 [17]
3 years ago
15

A solution of sodium hydroxide was titrated against a solution of sulfuric acid. How many moles of sodium hydroxide would react

with 1 mole of sulfuric acid?
Chemistry
1 answer:
jeka57 [31]3 years ago
5 0

Answer:

2 mole of Sodium hydroxide reacts with 1 mole of Sulfuric acid

Explanation:

Write down the equation in the beginning with reactants and products:

NaOH + H₂SO₄ → Na₂SO₄ + H₂0

Now try to balance it. Try with Na first:

2NaOH + H₂SO₄ → Na₂SO₄ + H₂0

Na atoms are balanced. There are 6 Oxygen atoms on the right and 5 on the left. Balance by increasing the H₂O moles:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂0

Check if H atoms are also balanced. They are. That means our final reaction is:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂0

2 Moles of NaOH reacts with 1 mole of H₂SO₄

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Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.210 M HClO(aq) with 0.210 M KOH(aq).
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<span>35.0 mL of 0.210 M
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 find moles of OH do the same thing for: 50.0 mL of 0.210 M HClO(aq) but for H+ they will cancel out: H+ + OH- -> H2O
 but you'll have some left over,
 pH=-log[H+] pOH
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     =14</span>
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An analytical chemist is titrating of a solution of nitrous acid with a solution of . The of nitrous acid is . Calculate the pH
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Answer:

pH = 2.69

Explanation:

The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>

<em />

The reaction of HNO₂ with KOH is:

HNO₂ + KOH → NO₂⁻ + H₂O + K⁺

Moles of HNO₂ and KOH that react are:

HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>

KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>

That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:

NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻

HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂

It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:

pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]

pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]

<h3>pH = 2.69</h3>
8 0
3 years ago
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