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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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5 0
3 years ago
Find the percent ionization of a 0.337 m hf solution. the ka for hf is 3.5 x 10-4. 1.1 % 1.2 x 10-2 % 3.2 % 3.5 x 10-2 % 4.7 %
Ugo [173]

To determine the percent ionization of the acid given, we make use of the acid equilibrium constant (Ka) given. It is the ration of the equilibrium concentrations of the dissociated ions and the acid. The dissociation reaction of the HF acid would be as follows:<span>

HF = H+ + F-

The acid equilibrum constant would be expressed as follows:

Ka = [H+][F-] / [HF] = 3.5 x 10-4

To determine the equilibrium concentrations we use the ICE table,
         HF             H+              F-
I      0.337           0                 0
C      -x              +x               +x
---------------------------------------------
E    0.337-x        x                   x 

3.5 x 10-4 = [H+][F-] / [HF] 
3.5 x 10-4 = [x][x] / [0.337-x] </span>

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x = 0.01069 = [H+] = [F-]

percent ionization = 0.01069 / 0.337 x 100  = 3.17%

8 0
3 years ago
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