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Doss [256]
3 years ago
14

Calculate the [H+]

Chemistry
1 answer:
ELEN [110]3 years ago
5 0

[H⁺]=6.696 x 10⁻⁵

pH = 4.174

<h3> Further explanation </h3>

Given

The concentration of 0.000295 M (2.95 x 10⁻⁴ M) butanoic acid solution

Required

the [H+]  and pH

Solution

Butanoic acid is the carboxylic acid group. Carboxylic acids are weak acids

For weak acid :

\tt [H^+]=\sqrt{Ka.M}

Input the value :

[H⁺]=√1.52 x 10⁻⁵ x 2.95 x 10⁻⁴

[H⁺]=6.696 x 10⁻⁵

pH = - log [H⁺]

pH = - log 6.696 x 10⁻⁵

pH = 5 - log 6.696

pH = 4.174

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1.33 dm3 of water at 70°C are saturated by 2.25
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Given that 4.50 dm³ of Pb(NO₃)₂ is cooled from 70 °C to 18 °C, the

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The volume of water 1.33 dm³ of water 70 °C.

The number of moles of Pb(NO₃)₂ that saturates 1.33 dm³ of water at 70 °C  = 2.25 moles

At 18 °C, the number of moles of Pb(NO₃)₂ that saturates 1.33 dm³ of water = 0.53 moles

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Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 70 °C is therefore;

1.33 dm³ contains 2.25 moles.

Number \ of \ moles \ in \ 4.50 \ dm^3 = \dfrac{2.25}{1.33} \times 4.50 \approx \mathbf{7.613 \, moles}

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 70 °C ≈ 7.613 moles

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 18 °C is therefore;

1.33 dm³ contains 0.53 moles

Number \ of \ moles \ in \ 4.50 \ dm^3 = \dfrac{0.53}{1.33} \times 4.50 \approx \mathbf{1.79 \, moles}

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 18 °C ≈ 1.79 moles

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Which gives;

Amount of solute deposited = 7.613 moles - 1.79 moles = 5.823 moles

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The molar mass of Pb(NO₃)₂ = 331 g/mol

The amount of solute deposited = Number of moles × Molar mass

Which gives;

The amount of solute deposited = 5.823 moles × 331 g/mol =<u> 1,927.413 g </u>

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