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ArbitrLikvidat [17]
3 years ago
10

80.0 mL of 0.30 M NaOH and 80.0 mL of 0.30 M HCl are mixed together. What is the approximate pH

Chemistry
1 answer:
Licemer1 [7]3 years ago
6 0

Answer:

  • <u><em>Neutral</em></u>

Explanation:

The number of moles of solute is equal to product of the molar concentration (molarity) and the volume (in liters) of solution.

Since the volumes and the molar concentrations of the<em> NaOH </em>and <em>HCl </em>solutions mixed are equal, each one of them contributes the same number of moles of solute.

Since every mol of NaOH produces one mol of OH⁻ ions and every mol of HCl produces one mol of H⁺ ion, the number of moles of OH ⁻ and H⁺ in solution are equal.

Thus, OH⁻ and H⁺ ions will be neutralized by the reaction:

  • OH⁻ (aq) + H⁺ (aq) ⇄ H₂O (l)

Which is strongly shifted to the right and has <em>neutral pH</em>.

Hence, you conclude that the approximate <em>pH of the solution is neutral.</em>

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The concentration of a dextrose solution prepared by diluting 14 ml of a 1.0 M dextrose solution to 25 ml using a 25 ml volumetric flask is 0.56M.

Concentration is defined as the number of moles of a solute present in the specific volume of a solution.

According to the dilution law, the degree of ionization increases on a dilution and it is inversely proportional to the square root of concentration. The degree of dissociation of an acid is directly proportional to the square root of a volume.

M₁V₁=M₂V₂

Where, M₁=1.0M, V₁=14ml, M₂=?, V₂=25ml

Rearrange the formula for M₂

M₂=(M₁V₁/V₂)

Plug all the values in the formula

M₂=(1.0M×14 ml/25 ml)

M₂=14 M/25

M₂=0.56 M

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The density of a saline (salt) solution is 1.27 g/mL. What is the mass of 326 milliliters of the solution?
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What is the gcf of 16 and 72? <br> a. 2 <br> b. 4 <br> c. 16 <br> d. 8
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Answer:

7. 3–ethyl–6 –methyldecane

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Explanation:

It is important to note that when naming organic compounds having two or more different substituent groups, we simply name them alphabetically.

The name of the compound given in the question above can be written as follow:

7. Obtaining the name of the compound.

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Naming alphabetically, we have

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