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GrogVix [38]
3 years ago
7

If 18.7ml of 0.01500M aqueous HCl is required to titrâtes 15.00ml of an aqueous solution of NaOH to the equivalence point, what

is the molarity of the NaOH solution
Chemistry
1 answer:
sweet [91]3 years ago
6 0

Answer:

0.0187 M

Explanation:

Step 1: Write the balanced neutralization reaction

NaOH + HCl ⇒ NaCl + H₂O

Step 2: Calculate the reacting moles of HCl

18.7 mL of 0.01500 M HCl react.

0.0187 L × 0.01500 mol/L = 2.81 × 10⁻⁴ mol

Step 3: Calculate the reacting moles of NaOH

The molar ratio of HCl to NaOH is 1:1. The reacting moles of NaOH are 1/1 × 2.81 × 10⁻⁴ mol = 2.81 × 10⁻⁴ mol.

Step 4: Calculate the molarity of NaOH

2.81 × 10⁻⁴ moles are in 15.00 mL of NaOH.

[NaOH] = 2.81 × 10⁻⁴ mol/0.01500 L = 0.0187 M

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<h3><u>Answer</u>;</h3>

1.0875 x 10-2 atm

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2O3(g) → 3O2(g)

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The average rate of disappearance of ozone ... is found to  

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This means (ignoring time)

∆[O3]/∆t = -7.25 × 10^–3 atm  

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Now use the other part of the expression:  

rate = +(1/3)∆[O2)∆t  

3.625 × 10–3 atm = +(1/3)∆[O2)/t  

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3 years ago
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The second option only.

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Sulfuric acid H₂SO₄ is the acid here. There are more than one classes of bases that can neutralize H₂SO₄. Among the options, there are:

Metal hydroxides

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Metal hydroxides react with sulfuric acid to produce water and the sulfate salt of the metal.

\text{Ca}(\text{OH})_{\bf 2}+\text{H}_2\text{SO}_4 \to \textbf{Ca}\textbf{SO}_{\bf 4} +{\bf 2}\;\text{H}_2\text{O}.

The formula for calcium sulfate \text{CaSO}_4 in option A is spelled incorrectly. Why? The charge on each calcium \text{Ca}^{2+} is +2. The charge on each sulfate ion {\text{SO}_4}^{2-} is -2. Unlike \text{Li}^{+} ions, it takes only one \text{Ca}^{2+} ion to balance the charge on each {\text{SO}_4}^{2-} ion. As a result, \text{Ca}^{2+} and {\text{SO}_4}^{2-} ions in calcium sulfate exist on a 1:1 ratio.

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The ion {\text{NH}_4}^{+} acts like a metal cation. Similarly to the metal hydroxides, NH₃ (or NH₄OH) neutralizes H₂SO₄ to produce water and a salt:

2\;\textbf{NH}_{\bf 4}\text{OH}+ \text{H}_2\text{SO}_4 \to (\textbf{NH}_{\bf 4})_2\text{SO}_4+2\;\text{H}_2\text{O}.

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