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3 years ago

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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?

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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<u>Explanation:</u>

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To calculate the hydroxide ion concentration, we first calculate pOH of the solution, which is:

pH + pOH = 14

We are given:

pH = 8.08

$pOH=14-8.08=5.92$

To calculate pOH of the solution, we use the equation:

$pOH=-\log[OH^-]$

Putting values in above equation, we get:

$5.92=-\log[OH^-]$

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The unknown salt given are formed by the combination of weak acid and strong acid which is NaOH

The chemical equation for the hydrolysis of $X^-$ ions follows:

$X^-(aq.)+H_2O(l)\rightleftharpoons HX(aq.)+OH^-(aq.);K_b$

<u>Initial:</u> 0.1

<u>At eqllm:</u> 0.1-x x x

Concentration of $OH^-=x=1.202\times 10^{-6}M$

The expression of $K_b$ for above equation follows:

$K_b=\frac{[OH^-][HX]}{[X^-]}$

Putting values in above expression, we get:

$K_b=\frac{(1.202\times 10^{-6})\times (1.202\times 10^{-6})}{(1-(1.202\times 10^{-6}))}\\\\K_b=1.445\times 10^{-11}M$

To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

$K_w=K_b\times K_a$

where,

$K_w$ = Ionic product of water = $10^{-14}$

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Putting values in above equation, we get:

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$K_a\text{ for HF}=6.8\times 10^{-6}$

$K_a\text{ for HCl}=1.3\times 10^{6}$

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So, the calculated $K_a$ is approximately equal to the $K_a$ of HF

Hence, the unknown salt is NaF

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