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3 years ago

-- A scientist combines 14 grams of potassium nitrate with an unknown amount of

Chemistry

1 answer:

igor_vitrenko [27]3 years ago

5 0

Answer:

The unknown amount of potassium chloride is 13.6 grams.

Explanation:

The reaction of 14 grams of KNO₃ with KCl produces a total mass of 27.6 grams of the products.

The law of conservation of mass tells us that the total mass of the reactants must be the same that the total mass of the products. So, we can find the mass of KCl as follows:

$m_{r} = m_{p}$

Where <em>r</em> is for reactants and <em>p </em>is for products

$m_{KNO_{3}} + m_{KCl} = m_{p}$

$14 g + m_{KCl} = 27.6 g$

$m_{KCl} = 27.6 g - 14 g = 13.6 g$

Therefore, the unknown amount of potassium chloride is 13.6 grams.

I hope it helps you!

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Answer:

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2 years ago

When neutralizing your acid extracts, you determined how much base solution was needed to fully neutralize your acid and reach a

alukav5142 [94]

Answer:

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Explanation:

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H⁺ + OH⁻ = H₂0

Hence, one mole of hydroxonium ion (H⁺) will combine with one mole of hydroxide ion (OH⁻) to give salt and water only.

In a completely neutralized reaction, the resulting salt is formed when there is complete dissociation of the acid and base to give salt and water with a pH of 7.

In the given question, the stated pH of between 8-9 tells us that the salt produced in this particular neutralization reaction is basic or alkaline. This usually occurs when a strong base reacts with a weak acid, producing a higher concentration of the hydroxide ion at equilibrium.

Hence the amount of base needed is the amount that would give one mole of the hydroxide ion needed to neutralise one mole of the hydroxonium ion from the acid.

If the concentration or molarity of the acid is known, then the exact amount of base required to neutralize it can be calculated. This is usually done via titrating the acid against drop wise solution of the base. Neutralization usually occurs when there is a change in colour of the resulting solution. The pH of the resulting solution can be determined using a litmus paper.

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3 years ago

What is the pressure of nitrogen in atmospheres of a sample that is at 745 mmHg?

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3 years ago

100.00 mL of 0.15 M nitrous acid (HNO2) are titrated with a 0.15 M NaOH solution. (a) Calculate the pH for the initial solution.

wolverine [178]

Answer:

a. pH = 2.04

b. pH = 3.85

c. pH = 8.06

d. pH = 11.56

Explanation:

The nitrous acid is a weak acid (Ka = 5.6x10⁻⁴) that reacts with NaOH as follows:

HNO₂ + NaOH → NaNO₂(aq) + H₂O(l)

a. At the beginning there is just a solution of 0.12M HNO₂. As Ka is:

Ka = [H⁺] [NO₂⁻] / [HNO₂]

Where [H⁺] and [NO₂⁻] ions comes from the same equilibrium ([H⁺] = [NO₂⁻] = X):

5.6x10⁻⁴ = X² / 0.15M

8.4x10⁻⁵ = X²

X = [H⁺] = 9.165x10⁻³M

As pH = -log [H⁺]

b. At this point we have HNO₂ and NaNO₂ (The weak acid and the conjugate base), a buffer. The pH of a buffer is obtained using H-H equation:

pH = pKa + log [NaNO₂] / [HNO₂]

<em>Where pH is the pH of the buffer,</em>

<em>And [NaNO₂] [HNO₂] could be taken as the moles of each compound.</em>

<em />

The initial moles of HNO₂ are:

0.100L * (0.15mol / L) = 0.015moles

The moles of base added are:

0.0800L * (0.15mol / L) = 0.012moles

The moles of base added = Moles of NaNO₂ produced = 0.012moles.

And the moles of HNO₂ that remains are:

0.015moles - 0.012moles = 0.003moles

Replacing in H-H equation:

pH = 3.25 + log [0.012moles] / [0.003moles]

c. At equivalence point all HNO2 reacts producing NaNO₂. The volume added of NaOH must be 100mL. That means the concentration of the NaNO₂ is:

0.15M / 2 = 0.075M

The NaNO₂ is in equilibrium with water as follows:

NaNO₂(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq) + Na⁺

The equilibrium constant, kb, is:

Kb = Kw/Ka = 1x10⁻¹⁴ / 5.6x10⁻⁴ = 1.79x10⁻¹¹ = [OH⁻] [HNO₂] / [NaNO₂]

<em>Where [OH⁻] = [HNO₂] = x</em>

<em />

1.79x10⁻¹¹ = [X] [X] / [0.075M]

1.34x10⁻¹² = X²

X = 1.16x10⁻⁶M = [OH⁻]

pOH = -log [OH-] = 5.94

pH = 14-pOH

d. At this point, 5mL of NaOH are added in excess, the moles are:

5mL = 5x10⁻³L * (0.15mol / L) =7.5x10⁻⁴moles NaOH

In 100mL + 105mL = 205mL = 0.205L. [NaOH] = 7.5x10⁻⁴moles NaOH / 0.205L =

3.66x10⁻³M = [OH⁻]

pOH = 2.44

pH = 14 - pOH

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2 years ago