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Lena [83]
4 years ago
13

When the NaOH dissolves, the temperature of the solution increases from 18.2 C to 24.5 C. Assuming the specific heat and density

of the solution is approximately that of water, what is the heat of solution of NaOH in kJ/mol?
Chemistry
1 answer:
ELEN [110]4 years ago
7 0

Answer:

The heat of solution is 1.05 kJ/mol

Explanation:

NaOH → Molar mass 40 g/m

This is the mass in 1 mol

Calorimetry formula:

Q = m . c . ΔT

ΔT = T° final - T° initial = 24.5°C - 18.2°C = 6.3°C

mass = 40 g

c = 4.186 kJ/kg°C (the same as water)

So we have to convert 40 g to kg

40 g/1000 = 0.04 kg

Q = 0.04 kg . 4.186 kJ/kg°C . 6.3 °C = 1.05 kJ

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Calculate the percent ionization of a 0.15 M benzoic acid solution in pure water and in a solution containing 0.10 M sodium benz
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Answer:

% ionization for benzoic acid = 0.08%

% ionization for sodium benzoate = 2.5%

The percentage ionization differ significantly because benzoic acid is a weak acid while sodium benzoate is a salt of benzoic acid. Their extent of dissociation also differ because they were compared in different solutions

Explanation:

Ka for pure water = 1.0 * 10-⁷

Ka for sodium benzoate = 6.5*10-⁵

1. For benzoic acid (C6H5COOH)

C6H5COOH ==== C6H5COO‐ + H+

0.15M 0 0

0.15-x x x

Ka = [C6H5COO-] [H+] / [C6H5COOH]

Ka = [X] [X] / 0.15 - X

1.0*10-⁷ = [X]² / 0.15 - x

But x is negligible compared to 0.15,

(1.0*10-⁷)*0.15 = x²

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X = 1.22 * 10-⁴

% ionization = ( [H+] / [C6H5COOH] ) * 100

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2. For C6H5COONa

Note: I will not repeat the same procedure of dissociation again since they're basically the same just the difference in ions

Ka for C6H5COONa = 6.5*10-⁵

6.5*10-⁵ = [X]² / (0.10 - X)

Cross-multiply both sides;

(6.5*10-⁵ * 0.10) = X²

Take square root of both side,

X= 2.5*10-³

% ionization = (2.5*10-³ / 0.10) *100

% ionization = 2.5%

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