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nikitadnepr [17]
3 years ago
11

Yall please help me bc there are literally no good answers to this with good votes or with lots of likes tht make sense on brain

ly so can u guys answer this PLZZZZ
on brainly for What charts, tables, or drawings would clearly show what you have learned in this lab?


Each chart, table, or drawing should have the following items:


A. An appropriate title


B. Appropriate labels
Chemistry
2 answers:
Sedaia [141]3 years ago
5 0

Answer:

B Appropriate labels

Explanation:

NemiM [27]3 years ago
3 0

Answer:

The answer is B i think, sorry if im wrong

Explanation:

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A buffer with a pH of 4.31 contains 0.31 M of sodium benzoate and 0.24 M of benzoic acid. What is the concentration of [ H 3 O ]
Mnenie [13.5K]

<u>Answer:</u> The hydronium ion concentration in the solution is 1.29\times 10^{-4}M

<u>Explanation:</u>

To calculate the molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles hydrochloric acid solution = 0.060 mol

Volume of solution = 1 L

Putting values in above equation, we get:

\text{Molarity of HCl}=\frac{0.060}{1L}\\\\\text{Molarity of HCl}=0.060M

The chemical reaction for aniline and HCl follows the equation:

                   C_6H_5COO^-+HCl\rightarrow C_6H_5COOH+Cl^-

<u>Initial:</u>           0.24          0.060              0.31

<u>Final:</u>             0.18          -                     0.37

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

pH=pK_a+\log(\frac{[\text{conjugate acid}]}{[\text{base}]})

pH=pK_a+\log(\frac{[C_6H_5COO^-]}{[C_6H_5COOH]})

We are given:

pK_a = negative logarithm of acid dissociation constant of benzoic acid = 4.2

[C_6H_5COO^-]=0.18M

[C_6H_5COOH]=0.37M

pH = ?

Putting values in equation 1, we get:

pH=4.2+\log(\frac{0.18}{0.37})\\\\pH=3.89

To calculate the hydronium ion concentration in the solution, we use the equation:

pH=-\log[H_3O^+]

pH = 3.89

Putting values in above equation, we get:

3.89=-\log[H_3O^+]

[H_3O^+]=10^{-3.89}=1.29\times 10^{-4}M

Hence, the hydronium ion concentration in the solution is 1.29\times 10^{-4}M

3 0
3 years ago
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