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3 years ago

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What is the pH of a solution prepared by mixing 25.00 mL of 0.10 M CH3CO2H with 25.00 mL of 0.010 M CH3CO2Na? Assume that the vo

lume of the solutions are additive and that K a = 1.8 × 10-5 for CH3CO2H.

1 answer:

denpristay [2]3 years ago

4 0

Answer:

pH = 3.74

Explanation:

<u>Given:</u>

Initial volume of CH3COOH, V1 = 25.00 ml

Initial concentration of CH3COOH, M1 = 0.10 M

Initial volume of CH3COONa, V1 = 25.00 ml

Initial concentration of CH3COONa, M2 = 0.010 M

Ka (CH3COOH) = 1.8*10^-5

<u>To determine:</u>

pH of the solution

<u>Calculation:</u>

The given solution of CH3COOH/CH3COONa is in fact a buffer whose pH is given by the Henderson-Hasselbalch equation where:

$pH = pKa + log\frac{[A-]}{[HA]} ----(1)$

where A- = concentration of conjugate base = [CH3COONa]

HA = weak acid = [CH3COOH]

Step 1: Calculate the final concentration of CH3COONa

V1 = 25.00 ml

V(final) = Total volume = 25.00 + 25.00 = 50.00 ml

M1 = 0.010 M

$M1V1 = M2V2\\\\M2 = \frac{M1V1}{V2} = \frac{0.010 M * 25.00 ml}{50.00ml} =0.005M$

<u>Step 2</u>: Calculate the final concentration of CH3COOH

V1 = 25.00 ml

V(final) = Total volume = 25.00 + 25.00 = 50.00 ml

M1 = 0.10 M

$M1V1 = M2V2\\\\M2 = \frac{M1V1}{V2} = \frac{0.10 M * 25.00 ml}{50.00ml} =0.05M$

<u>Step 3</u>: Calculate the pH

Based on equation (1)

$pH = pKa + log\frac{[CH3COONa]}{[CH3COOH]} ----(1)$

pKa = -log Ka = -log(1.8*10^-5) = 4.74

$pH = -logKa + log\frac{[CH3COONa]}{[CH3COOH]}$

$pH = 4.74 + log\frac{[0.005]}{[0.05]}$

pH = 3.74

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The initial volume of neon gas is $40.81 \text{ m}^{3}$ .

The final volume of neon gas is $50.00 \text{ m}^{3}$ .

The initial temperature value is $23.5 \text{ } ^{\circ} \text{C}$ .

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The final temperature

Given Condition:

The pressure is constant.

Mass of gas is fixed.

Solution:

Step 1: Modify the mathematical expression for Charles’ law for two different temperature and volume values as follows:

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$V_{1}$ is the initial volume of the gas.

$V_{2}$ is the final volume of the gas.

$T_{1}$ is the initial temperature of the gas.

$T_{2}$ is the final temperature of the gas.

Step 2: Rearrange equation (2) for .

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The conversion factor to convert degree Celsius to Kelvin is,

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Substitute $23.5\text{ }^{\circ} \text{C}$ for $T(^{\circ}\text{C})$ in equation (3) to convert temperature from degree Celsius to Kelvin.

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Important note:

The temperature must be in Kelvin.

The condition of fixed mass and fixed pressure must be fulfilled in order to apply Charles’ law.

Learn More:

1. Gas laws brainly.com/question/1403211

2. Application of Charles’ law brainly.com/question/7434588

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: States of matter

Keywords: neon, volume, occupies, temperature, Kelvin, degree Celsius, Charle’s law, constant pressure, fixed mass, 40.81 m^3 , 50.00 m^3 , 23.5 degree C , celsius , 363 K , sates of matter, initial volume, final volume, initial temperature, final temperature, V1 , V2 , T1 , T2 .

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