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Anni [7]
3 years ago
15

25 mL of 0.10 M aqueous acetic acid is titrated with 0.10 M NaOH(aq). What is the pH after 30 mL of NaOH have been added?

Chemistry
1 answer:
weeeeeb [17]3 years ago
6 0

Answer:

pH = 11.95≈12

Explanation:

Remember  the reaction among aqueous acetic acid (CH_3COOH) and aqueous sodium hydroxide (NaOH)

CH_3COOH + NaOH ->  CH_3COONa + H_2O

First step. Need to know how much moles of the substances are present

0.1 \frac{mol NaOH}{L} * 0.030 L = 0.003 mol NaOH[tex]0.1 \frac{mol CH_3COOH}{L} * 0.025 L= 0.0025 mol CH_3COOHSecond step. Know wich substance is in excess.0.0025 mol CH_3COOH * [tex]1 mol NaOH/ 1 mol CH_3COOH = 0.0025 mol NaOH

0.003 mol NaOH * 1 mol CH_3COOH/ 1 mol NaOH = 0.003 mol CH_3COOH[/tex]

NaOH is in excess. Now, how much?

0.003 mole NaOH - 0.0025 mole NaOH = 0.0005 mole NaOH

Then, that amount in excess would be responsable for the pH.

Third step. Know the pH

Remember that pH= -log[H+]

According to the dissociation of water equilibrium

Kw=[H+]*[OH-]= 10^(-14)

The dissociation of NaOH is

NaOH -> Na^{+} + OH^{-}

Now, concentration of OH^{-}[/tex] would be given for the excess of NaOH.

[OH-]= 0.0005 mole / 0.055 L = 0.00909 M

Careful: we have to use the total volumen

Les us to calculate pH

pH= -log [H+]\\pH= -log \frac{K_w}{[OH-]} \\pH= 11.95

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<h3>The answer is 9.0 kg/L</h3>

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I need to check some chemistry questions. Help with any question is appreciated! :) Please include an explanation with your conc
finlep [7]

1. 4.67 kg; 2. 4.8 ×10^5 kg; 3. 0.106 cm^3; 4. 1.7 g/cm^3

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<em>Q2. Mass of Pb </em>

<em>Step 1</em>. Calculate the <em>volume of the Pb</em>.

<em>V = lwh</em> = 6.0 m × 3.5 m × 2.0 m = 42.0 m^3

<em>Step 2</em>. Calculate the <em>mass of the Pb</em>.

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<em>Q3. Volume of displaced water </em>

Volume of Ag = 0.987 g × (1 cm^3/9.320 g) = 0.106 cm^3

<em>Archimedes</em>: volume of displaced water = volume of Ag = <em>0.106 cm^3</em>

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<em>Step 1</em>. Convert <em>ounces to grams </em>

Mass = 3.35 oz × (28.35 g/1 oz) = 94.97 g

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<em>V = lwh</em> = 3.0 in × 2.5 in × 0.45 in = 3.38 in^3

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<em>V</em> = 3.38 in^3 × (2.54 cm/1 in)^3 = 55.3 cm^3

<em>Step 4</em>. Calculate the <em>density</em>

ρ = <em>m</em>/<em>V</em> = (94.97 g/55.3 cm^3) = 1.7 g/cm^3 (magnesium?)

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