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1 year ago

3. ml of 1.00 M LiOH to 500.00 ml of 0.075 M of LiOH (Dilution Problem)

Chemistry

1 answer:

castortr0y [4]1 year ago

6 0

The volume (in mL) of the 1.00 M LiOH needed to prepare 500.00 mL of 0.075 M of LiOH is 37.5 mL

<h3>How do I determine the volume (in mL) of LiOH needed?</h3>

Dilution formula is given as follow:

M₁V₁ = M₂V₂

Where

M₁ is the molarity of stock solution
V₁ is the volume of stock solution
M₂ is the molarity of diluted solution
V₂ is the volume of diluted solution

Applying the above formula, we obtain the volume (in mL) of LiOH needed to prepare 500.00 mL of 0.075 M of LiOH. This is illustrated below:

Molarity of stock solution of LiOH (M₁) = 1.00 M
Volume of diluted solution of LiOH (V₂) = 500.00 mL
Molarity of diluted solution of LiOH (M₂) = 0.075 M
Volume of stock solution of LiOH needed (V₁) =?

M₁V₁ = M₂V₂

1 × V₁ = 0.075 × 500

V₁ = 37.5 mL

Thus, the volume (in mL) of LiOH needed is 37.5 mL

Learn more about volume:

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Explanation:

<em>The half-reactions are missing, but I will propose some to show you the general procedure and then you can apply it to your equations.</em>

<em>Suppose we have the following half-reactions.</em>

<em />

To identify how to make a spontaneous cell, we need to consider the standard reduction potentials (E°red). The half-reaction with the higher E°red will occur as a reduction (in the cathode), whereas the one with the lower E°red will occur as an oxidation (in the anode).

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Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)

The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

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3. ml of 1.00 M LiOH to 500.00 ml of 0.075 M of LiOH (Dilution Problem)​

3. ml of 1.00 M LiOH to 500.00 ml of 0.075 M of LiOH (Dilution Problem)