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3 years ago

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Phenolphthalein is an indicator that turns from colorless (acidic form) to magenta (basic form) and has a pKa of 9.40. What is t

he ratio of the magenta phenolphthalein concentration to the colorless phenolphthalein concentration ([magenta phenolphthalein]/[colorless phenolphthalein]) at a pH of 11?

1 answer:

frosja888 [35]3 years ago

7 0

Answer:

40:1 is the ratio of the magenta phenolphthalein concentration to the colorless phenolphthalein concentration.

Explanation:

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[salt]}{[acid]})$

$pH=pK_a+\log(\frac{[magenta(Php)]}{[Php]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of phenolphthalein = 9.40

$[magenta(Php)]$ = concentration of magenta phenolphthalein

$[Php]$ = concentration of colorless phenolphthalein

pH = 11

Putting values in above equation, we get:

$11=9.40+\log(\frac{[magenta(Php)]}{[Php]})$

$\log(\frac{[magenta(Php)]}{[Php]})=11-9.40=1.6$

$\frac{[magenta(Php)]}{[Php]}=10^{1.6}=39.81 :1 \approx 40:1$

40:1 is the ratio of the magenta phenolphthalein concentration to the colorless phenolphthalein concentration.

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$\text{Concentration of }FeSCN^{2+}=\frac{0.0020mmol}{9.0mL+1.0mL}=0.00020M$

Using Beer-Lambert's law :

$A=\epsilon \times C\times l$

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A = absorbance of solution

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$\epsilon$ = molar absorptivity coefficient

$\epsilon$ and l are same for stock solution and dilute solution. So,

$\epsilon l=\frac{A}{C}=\frac{0.480}{0.00020M}=2400M^{-1}$

For trial solution:

The equilibrium concentration of $SCN^-$ is,

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

$[SCN^-]_{initial}$ = 0.00050 M

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Now calculate the concentration of $SCN^-$ .

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$[SCN^-]_{eqm}=(0.00050M)-(9.17\times 10^{-5}M)$

$[SCN^-]_{eqm}=4.58\times 10^{-8}M$

Therefore, the equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

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