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bija089 [108]
3 years ago
13

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?
Chemistry
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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For each ionic compound formula, identify the main group to which X belongs: (c) X2SO4.
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The answer is group 1A.

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Group 1A elements form X⁺ type of ions because they are metals. So, they have the tendency to lose electrons.

Hence, X must belong to the 1A group.

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6 0
1 year ago
What is the percent yield of a reaction in which 51.5 g of tungsten(VI) oxide (WO3) reacts with excess hydrogen gas to produce m
Rus_ich [418]

Answer:

The percent yield of a reaction is 48.05%.

Explanation:

WO_3+3H_2\rightarrow W+3H_2O

Volume of water obtained from the reaction , V= 5.76 mL

Mass of water = m = Experimental yield of water

Density of water = d = 1.00 g/mL

M=d\times V = 1.00 g/mL\times 5.76 mL=5.76 g

Theoretical yield of water : T

Moles of tungsten(VI) oxide = \frac{51.5 g}{232 g/mol}=0.2220 mol

According to recation 1 mole of tungsten(VI) oxide gives 3 moles of water, then 0.2220 moles of tungsten(VI) oxide will give:

\frac{3}{1}\times 0.2220 mol=0.6660 mol

Mass of 0.6660 moles of water:

0.666 mol × 18 g/mol = 11.988 g

Theoretical yield of water : T = 11.988 g

To calculate the percentage yield of reaction , we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

=\frac{m}{T}\times 100=\frac{5.76 g}{11.988 g}\times 100=48.05\%

The percent yield of a reaction is 48.05%.

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