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Oksana_A [137]
3 years ago
15

Standardization of a Borax solution (Na2B4O7). You are given a 1.044 M solution of H2SO4. It takes 2.63 mL of this H2SO4 to reac

h the end point. Knowing it takes 1 H2SO4 to neutralize 2 Na2B4O7, what was the concentration of this Borax solution?
Chemistry
1 answer:
strojnjashka [21]3 years ago
8 0

The question is incomplete, the complete question is:

Standardization of a Borax solution (Na2B4O7). A student titrates a 20.00 mL sample of an aqueous borax solution with 1.044 M H2SO4. It takes 2.63 mL of acid to reach the equivalence point. Knowing it takes 1 H2SO4 to neutralize 2 Na2B4O7, what was the concentration of this Borax solution?

<u>Answer: </u>The concentration of borax solution is 0.069 M.

<u>Explanation:</u>

To calculate the concentration of borax solution, the formula used is:

n_1C_1V_1=n_2C_2V_2 ....(1)

where,

n_1, C_1\text{ and }V_1 are the n-factor, concentration and volume of sulfuric acid

n_2,C_2\text{ and }V_2 are the n-factor, concentration and volume of borax solution.

We are given:

n_1=1\\C_1=1.044M\\V_1=2.63mL\\n_2=2\\C_2=?M\\V_2=20mL

Putting values in equation 1, we get:

1\times 1.044\times 2.63=2\times C_2\times 20\\\\C_2=\frac{1\times 1.044\times 2.63}{2\times 20}\\\\C_2=0.069M

Hence, the concentration of borax solution is 0.069 M.

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Calculate the pH and fraction of dissociation ( α ) for each of the acetic acid ( CH 3 COOH , p K a = 4.756 ) solutions. A 0.002
marysya [2.9K]

Answer:

The degree of dissociation of acetic acid is 0.08448.

The pH of the solution is 3.72.

Explanation:

The pK_a=4.756

The value of the dissociation constant = K_a

pK_a=-\log[K_a]

K_a=10^{-4.756}=1.754\times 10^{-5}

Initial concentration of the acetic acid = [HAc] =c = 0.00225

Degree of dissociation = α

HAc\rightleftharpoons H^++Ac^-

Initially

c

At equilibrium ;

(c-cα)                                cα        cα

The expression of dissociation constant is given as:

K_a=\frac{[H^+][Ac^-]}{[HAc]}

1.754\times 10^{-5}=\frac{c\times \alpha \times c\times \alpha}{(c-c\alpha)}

1.754\times 10^{-5}=\frac{c\alpha ^2}{(1-\alpha)}

1.754\times 10^{-5}=\frac{0.00225 \alpha ^2}{(1-\alpha)}

Solving for α:

α = 0.08448

The degree of dissociation of acetic acid is 0.08448.

[H^+]=c\alpha = 0.00225M\times 0.08448=0.0001901 M

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3 0
3 years ago
What is the percent composition by mass of hydrogen in nh4hco3
Simora [160]

<u>Answer:</u> The percent composition by mass of hydrogen in given compound is 6.33 %

<u>Explanation:</u>

We are given:

A chemical compound having chemical formula of NH_4HCO_3

It is made up by the combination of 1 nitrogen atom, 5 hydrogen atoms, 1 carbon atom and 3 oxygen atoms

To calculate the percentage composition by mass of hydrogen in the compound, we use the equation:

\%\text{ composition of Hydrogen}=\frac{\text{Mass of hydrogen}}{\text{Mass of compound}}\times 100

Mass of compound = [(1\times 14)+(5\times 1)+(1\times 12)+(3\times 16)]=79g/mol

Mass of hydrogen = (5\times 1)=5g/mol

Putting values in above equation, we get:

\%\text{ composition of Hydrogen}=\frac{5g/mol}{79g/mol}\times 100=6.33\%

Hence, the percent composition by mass of hydrogen in given compound is 6.33 %

6 0
4 years ago
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matrenka [14]

Answer:

Yes.

Explanation:

Yes, this difference of readings will definitely affect the results of the experiment as well as the E values because the readings taken by both students are different from one another. There is a fault in one of the thermometer because both shows different readings of temperature of the same solution. This will affect the overall experiment and due to this error, we are unable to tell that which one reading is correct so the answer is uncertain or unsure.

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What is the volume of a sample of CO2 at STP that has a volume of 75.0mL at 30.0°C and 91kPa
Nezavi [6.7K]

60.7 ml is the volume of a sample of CO2 at STP that has a volume of 75.0mL at 30.0°C and 91kPa.

Explanation:

Data given:

V1 = 75 ml

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P1 = 91 KPa

V2  =?

P2 = 1 atm or 101.3 KPa

T2 = 273.15 K

At STP the pressure is 1 atm and the temperature is 273.15 K

applying Gas Law:

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putting the values in the equation of Gas Law:

V2 = \frac{P1V1T2}{P2T1}

V2 = \frac{91 X 75 X 273.15}{303.15 X 101.3}

V2 = 60.7 ml

at STP the volume of carbon dioxide gas is 60.7 ml.

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