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Ad libitum [116K]
3 years ago
10

How would the calculated concentration of the hci be affected if the sodium hydroxide were poured from ma beaker that contained

some water before the naoh were added to it
Chemistry
1 answer:
Mekhanik [1.2K]3 years ago
5 0

Answer:

This question is incomplete

Explanation:

This question is incomplete. However, the beaker that contained some water before NaOH were added means that the resulting solution in that beaker will be more dilute. When this diluted sodium hydroxide solution is added to HCl (not hci), the reaction below occurs

HCl + NaOH ⇒ NaCl + H₂O

The reaction above is a neutralization reaction. <u>The concentration of the acid (HCl) will reduce when a base (sodium hydroxide) is added and will also reduce more because of the presence of more water (in the base) which normally reduces the concentration of ions present in an acid or a base to become more dilute.</u>

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You need to prepare .200M solution of hydrochloric acid (HCl). If you took out .830ml of the 12.0M stock solution. How much wate
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Answer:

You must add 48.97 mL of water to make the 0.200 M diluted solution.

Explanation:

In chemistry, dilution is the reduction in concentration of a chemical in a solution. In other words, it is the process of reducing the concentration of solute in solution, simply adding more solvent to the solution.

In a dilution, the quantity or mass of the solute is not changed but only that of the solvent. As only solvent is being added, by not increasing the amount of solute the concentration of the solute decreases.

The expression for the dilution calculations is:

Cinitial* Vinitial = Cfinal* Vfinal

In this case:

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Replacing:

12 M*0.830 mL= 0.200 M*Vfinal

Solving:

Vfinal=\frac{12 M*0.830 mL}{0.200 M}

Vfinal= 49.8 mL

Since 0.830 mL is the volume you initially have of HCl, the amount of water you must add is:

49.8 mL - 0.830 mL= 48.97 mL

<u><em>You must add 48.97 mL of water to make the 0.200 M diluted solution.</em></u>

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