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Contact [7]
3 years ago
14

The forensic technician at a crime scene has just prepared a luminol stock solution by adding 16.0 g of luminol to H2O creating

a solution with a total volume of 75.0 mL. What is the molarity of the stock solution of luminol?
Chemistry
1 answer:
san4es73 [151]3 years ago
4 0

Answer : The concentration of luminol in the solution is, 1.20 M.

Explanation : Given,

Mass of luminol = 16.0 g

Volume of solution = 75.0 mL

Molar mass of luminol = 177.16 g/mole

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

\text{Molarity}=\frac{\text{Mass of luminol}\times 1000}{\text{Molar mass of luminol}\times \text{Volume of solution (in mL)}}

Now put all the given values in this formula, we get:

\text{Molarity}=\frac{16.0g\times 1000}{177.16g/mole\times 75.0mL}=1.20mole/L=1.20M

Therefore, the concentration of luminol in the solution is, 1.20 M.

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Since acids have 1 more proton (H+ - ions) than base, and the acid gives it away, doesn't that mean that they switch roles? Acid
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Answer:

In an acid-base equilibrium, acid becomes a conjugate base and base becomes a conjugate acid.

Explanation:

Let's remember the Bronsted-Lowry theory to answer this specific question. According to the theory, acid is a proton donor, while a base is a proton acceptor.

Consider an acid in a form HA (aq) and base in a form of B (aq). Since acid is a proton donor, it will donate its hydrogen ion to the base, B. The resultant products would be A^{-} (aq) and BH^{+} (aq).

Remember that an acid-base reaction is an equilibrium reaction. This means we may also look at this proton transfer reaction from the product side towards the reactants. Summarizing what has been said, we may write the equilibrium as:

HA (aq) + B (aq) ⇄ BH^{+} (aq) + A^{-} (aq)

Now acid, HA, donates a proton to become a conjugate base. The conjugate base, if we look from the reverse equation side, is actually a base, since it can accept a proton to become HA. Similarly, B accepts a proton to become a conjugate acid. Looking from the reverse reaction, it can now donate a proton, so in reality we can consider it a base.

To summarize, your logic is correct.

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How do you think life on Earth would be different without the evolution of angiosperms? Provide two examples, each with an expla
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2 years ago
A pH of 2 indicates a(n) _____.
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A pH of 2 indicates a acid
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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⁺]

<h3>pH = 2.04</h3><h3 />

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>pKa is -log Ka = 3.25</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]

<h3>pH = 3.85</h3><h3 />

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>[NaNO₂] = 0.075M</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

<h3>pH = 8.06</h3><h3 />

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

<h3>pH = 11.56</h3>
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