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sleet_krkn [62]
3 years ago
13

A 25.0-mL sample of 0.150 M butanoic acid is titrated with a 0.150 M NaOH solution. What is the pH before any base is added? The

Ka of butanoic acid is 1.5 × 10-5.
Chemistry
1 answer:
hoa [83]3 years ago
7 0
HA ⇄ H⁺ + A⁻ 
so:
\frac{[H^+][A^-]}{[HA]} = 1.5 x 10^{-5}
and now:
\frac{(x)(x)}{(0.150-x)} = 1.5 x 10⁻⁵
x is considered very small compared to 0.15
x² = 2.25 x 10⁻⁶
x = 1.5 x 10⁻³
So [H⁺] = 1.5 x 10⁻³
pH = - log [H⁺] = - log (1.5 x 10⁻³) = 2.83
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A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and
sammy [17]

<u>Answer:</u> The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

Where,

m_{solute} = Given mass of solute (S_8) = 0.800 g

M_{solute} = Molar mass of solute (S-8) = 256.52 g/mol

W_{solvent} = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m

  • <u>Calculation for freezing point of solution:</u>

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

or,

\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

K_f = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC

Hence, the freezing point of solution is 16.5°C

  • <u>Calculation for boiling point of solution:</u>

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

or,

\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

K_f = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC

Hence, the boiling point of solution is 118.2°C

5 0
3 years ago
How many grams of aluminum oxide (Al2O3) will be formed from 10.0 grams of aluminum (Al)? 4 AL + 3 02 --&gt; 2 AL203​
d1i1m1o1n [39]

Answer:

Mass = 18.9 g

Explanation:

Given data:

Mass of Al₂O₃ formed = ?

Mass of Al = 10.0 g

Solution:

Chemical equation:

4Al + 3O₂      →       2Al₂O₃

Number of moles of Al:

Number of moles = mass/molar mass

Number of moles = 10.0 g/ 27 g/mol

Number of moles = 0.37 mol

Now we will compare the moles of Al and Al₂O₃.

                      Al          :          Al₂O₃

                       4           :            2

                     0.37        :         2/4×0.37 = 0.185 mol

Mass of Al₂O₃:

Mass = number of moles × molar mass

Mass = 0.185 mol × 101.9 g/mol

Mass = 18.9 g

4 0
2 years ago
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frez [133]

Answer:

Lithium

Explanation:

I believe it is Lithium. It can easily be cut with a knife and is the 3rd lightest non gas on the periodic table that is not a gas.

7 0
2 years ago
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galina1969 [7]
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2 years ago
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4 0
2 years ago
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