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Murrr4er [49]
3 years ago
9

You have separate solutions of hcl and h2so4 with the same concentrations in terms of molarity. you wish to neutralize a solutio

n of naoh. which acid solution would require more volume (in ml) to neutralize the base?
Chemistry
2 answers:
Vera_Pavlovna [14]3 years ago
8 0
HCl + NaOH = NaCl + H₂O

H₂SO₄ + 2NaOH = Na₂SO₄ + 2H₂O ⇒ <span>0.5H₂SO₄ + NaOH = 0.5Na₂SO₄ + H₂O

V(HCl)=2V(H₂SO₄)

</span><span>The volume of the hydrochloric acid solution (HCl) is twice as large as the volume of the sulfuric acid solution (H</span>₂SO₄)<span>.
</span>
zlopas [31]3 years ago
4 0

Answer:

The sodium hydroxide under equal concentrations

Explanation:

Hello,

At first, lets consider the neutralization reactions for the involved acids with sodium hydroxide:

NaOH+HCl-->NaCl+H_2O\\2NaOH+H_2SO_4-->Na_2SO_4+2H_2O

As you can see the mole ration between hydrochloric acid and sodium hydroxide is 1 to 1, it means that for equal concentations, the volume of acid will equal the volume of base. However, for sulfuric acid we have a 2 to 1 ratio (2 moles of hydroxide per 1 mole of acid), it means that for equal concentrations, the half of sulfuric acid's volume will be required to neutralize the sodium hydroxide. In such a way, the hydrochloric acid's solution will require more volume to neutralize the sodium hydroxide under equal concentrations.

Best regards.

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Will give lots of points if answered correctly. Determine the kb for chloroform when 0.793 moles of solute in 0.758 kg changes t
Liono4ka [1.6K]

Answer: The value of K_{b} for chloroform is 3.62^{o}C/m when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

Explanation:

Given: Moles of solute = 0.793 mol

Mass of solvent = 0.758

\Delta T_{b} = 3.80^{o}C

As molality is the number of moles of solute present in kg of solvent. Hence, molality of given solution is calculated as follows.

Molality = \frac{no. of moles}{mass of solvent (in kg)}\\= \frac{0.793 mol}{0.758 kg}\\= 1.05 m

Now, the values of K_b is calculated as follows.

\Delta T_{b} = i\times K_{b} \times m

where,

i = Van't Hoff factor = 1 (for chloroform)

m = molality

K_{b} = molal boiling point elevation constant

Substitute the values into above formula as follows.

\Delta T_{b} = i\times K_{b} \times m\\3.80^{o}C = 1 \times K_{b} \times 1.05 m\\K_{b} = 3.62^{o}C/m

Thus, we can conclude that the value of K_{b} for chloroform is 3.62^{o}C/m when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

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