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Rudik [331]
3 years ago
5

Write out the balanced chemical equation and the net ionic equation for the reaction between hydrochloric acid and sodium hydrox

ide, using smallest whole number stoichiometric coefficients.
Chemistry
1 answer:
polet [3.4K]3 years ago
7 0

Answer:

HCl(aq) + NaOH(aq) ⇒ NaCl(aq) + H₂O(l)

H⁺(aq) + OH⁻ ⇒ H₂O(l)

Explanation:

Let's consider the balanced chemical equation (molecular equation) for the reaction between hydrochloric acid and sodium hydroxide. This is a neutralization reaction.

HCl(aq) + NaOH(aq) ⇒ NaCl(aq) + H₂O(l)

The complete ionic equation includes all the ions and the molecular species.

H⁺(aq) + Cl⁻(aq) + Na⁺(aq) + OH⁻ ⇒ Na⁺(aq) + Cl⁻(aq) + H₂O(l)

The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and the molecular species.

H⁺(aq) + OH⁻ ⇒ H₂O(l)

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Explanation:

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3 years ago
Calculate the volume of 1.0 x 10-4 M CV+ solution that needs to be added to a 25.0 mL volumetric flask and diluted with deionize
xenn [34]

Answer:

The answers are: 2.5 ml (first part) and 3.0 ml (second part)

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In order to calculate volumes required to prepare diluted solutions we use the following equation:

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Where Vc and Cc are the volume and concentration respectively of concentrated solution (higher concentration) whereas Vd and Cd are volume and concentration of diluted solution (lower concentration).

In both problems we want to prepare a diluted solution and we know the final concentration (Cd) and final volume (Vd) and the initial concentration (Cc).

In first part, we have: Cc= 1.0 10⁻⁴ M; Vd= 25 ml; Cd= 1.0 10⁻⁵ M

Vc= Vd x Cd / Cc= (25 ml x 1.0 10⁻⁵ M)/1.0 10⁻⁴ M = 2.5 ml

Notice that Cc/Cd= 1.0 10⁻⁴ M/Cd= 1.0 10⁻⁵ M= 10 (so, we have to dilute the solution 10 times, and for this we have to take a volume 10 times lower than the final volume).

To prepare the solution, we take 2.5 ml of 1.0 10⁻⁵ M CV+, we dispense the volume in a 25 ml volumetric flask and then we add water until complete 25 ml (aproximately 22.5 ml of water).

In the second part is the same. We have: Vd= 10 ml; Cc= 1.0 10⁻⁴M; Cd= 3.0 10⁻⁵M.

Vc= Vd x Cd / Cc= (10 ml x 3.0 10⁻⁵ M)/1.0 10⁻⁴ M = 3 ml

To prepare the solution, we take 3 ml of 3.0 10⁻⁵ M CV+, we dispense the volume in a 10 ml volumetric flask and then we add water until we complete 10 ml (aproximately 7 ml of water).

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