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3 years ago

Complete the acid-base reaction between butyric acid HC4H7O2 and potassium hydroxide KOH.

1 answer:

6 0

An acid-base reaction or a neutralization reaction is a chemical reaction that occurs between an acid and a base producing a salt and water. The acids and bases can be strong or weak depending on their degree of ionization in water.

Butyric acid is a weak acid and in water it is ionized in the following way, loosing a proton (H+):

HC4H7O2 (aq) ⇆ H+ (aq) + C4H7O2- (aq)

On the other hand, potassium hydroxide is a strong base, so it will be completely ionized in water:

KOH(aq) → K+(aq) + OH-(aq)

Then the net acid-base reaction between butyric acid and KOH is:

HC4H7O2 (aq) + OH- (aq) ⇆ H2O + C4H7O2- (aq)

It is valid to consider only the OH- produced from the ionization of KOH in water since, as mentioned, this molecule is completely ionized. Also, we do not include the K + in the net equation since it is a spectator ion, it does not undergo chemical changes.

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3 years ago

what is the percent by mass of a solution that contains 30 grams of potassium nitrite in 0.5 kilograms of water?

Makovka662 [10]

Answer:

5.66 %.

Explanation:

mass percent is the ratio of the mass of the solute to the mass of the solution multiplied by 100.

mass % = (mass of solute/mass of solution) x 100.

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mass of the solution = mass of water + mass of potassium nitrite = 500.0 g + 30.0 g = 530.0 g.

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3 years ago

Oh, no! You just spilled 85.00 mL of 1.500 M sulfuric acid on your lab bench and need to clean it up immediately! Right next to

vredina [299]

Explanation:

We will balance equation which describes the reaction between sulfuric acid and sodium bicarbonate: as follows.

$H_2SO_4(aq) + 2NaHCO_3(s) \rightarrow Na2SO_4(aq) + 2H_2O(l) + 2CO_2(g)$

Next we will calculate how many moles of $H_2SO_4$ are present in 85.00 mL of 1.500 M sulfuric acid.

As, Molarity = $\frac{\text{moles of solute}}{\text{liters of solution
}}$

1.500 M = $\frac{n}{0.08500 L
}$

n = 0.1275 mol $H_2SO_4$

Now set up and solve a stoichiometric conversion from moles of $H_2SO_4$ to grams of $NaHCO_3$ . As, the molar mass of $NaHCO_3$ is 84.01 g/mol.

$0.1275 mol H_2SO_4 \times (\frac{2 mol NaHCO_3}{1 mol H_2SO_4}) \times (\frac{84.01 g NaHCO_3}{1 mol NaHCO_3})$

= 21.42 g $NaHCO_3$

So unfortunately, 15.00 grams of sodium bicarbonate will "not" be sufficient to completely neutralize the acid. You would need an additional 6.42 grams to complete the task.

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3 years ago

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masha68 [24]

Answer:

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3 years ago

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