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rusak2 [61]
3 years ago
13

What is the pH of a buffer that consists of 0.254 M CH3CH2COONa and 0.329 M CH3CH2COOH? Ka of propanoic acid, CH3CH2COOH is 1.3

x 10-5 Enter your answer with two decimal places.
Chemistry
1 answer:
Shtirlitz [24]3 years ago
7 0

Answer:

4.77 is the pH of the given buffer .

Explanation:

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

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

pH=-\log[K_a]+\log(\frac{[salt]}{[acid]})

pH=-\log[K_a]+\log(\frac{[CH_3CH_2COONa]}{[CH_3CH_2COOH]})

We are given:

K_a = Dissociation constant of propanoic acid = 1.3\times 10^{-5}

[CH_3CH_2COONa]=0.254 M

[CH_3CH_2COOH]=0.329 M

pH = ?

Putting values in above equation, we get:

pH=-\log[1.3\times 10^{-5}]+\log(\frac{[0.254 M]}{[0.329]})

pH = 4.77

4.77 is the pH of the given buffer .

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Using the law of constant proportions  which says that within the same compound, elements exist in fixed ratios. 

Therefore; we can use the ratio of total mass to the mass of carbon, to determine the amount of carbon in another sample.

Mass C / Mass CH4 = Mass C / Mass CH4

43.2 g / 57.6 g = Mass C / 37.8 g

Mass C = 37.8 g × 43.2 g / 57.6 g 

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Hence; the percentage of carbon will be; 

=(28.35/ 37.8 )× 100%

= 75 % 

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4 years ago
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Answer:

B

Explanation:

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Rzqust [24]

Answer:

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The density of benzene is 0.879g/mL. The volume (with 1 decimal) of 131.9g sample of benzene is how many mL?
timurjin [86]

Answer:

150.1 mL

Explanation:

Step 1: Given data

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Step 2: Calculate the volume of the sample of benzene

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