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3 years ago

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A chemistry student needs to standardize a fresh solution of sodium hydroxide. She carefully weighs out 385.mg of oxalic acid H2

C2O4, a diprotic acid that can be purchased inexpensively in high purity and dissolves it in 250.mL of distilled water. The student then titrates the oxalic acid solution with her sodium hydroxide solution. When the titration reaches the equivalence point, the student finds she has used 123.7mL of sodium hydroxide solution. Calculate the molarity of the student's sodium hydroxide solution.

1 answer:

Vlada [557]3 years ago

4 0

Answer:

The molarity of the sodium hydroxide solution is 0.0692 M

Explanation:

<u>Step 1: </u>Data given

Mass of H2C2O4 = 385 mg = 0.385 grams

volume = 250 mL = 0.250 L

Volume of NaOH = 123.7 mL = 0.1237 L

Molar mass H2C2O4 = 90.03 g/mol

<u>Step 2</u>: The balanced equation

2NaOH + H2C2O4 → Na2C2O4 + 2H2O

<u>Step 3:</u> Calculate moles H2C2O4

Moles H2C2O4 = mass H2C2O4 / molar mass H2C2O4

Moles H2C2O4 = 0.385 grams / 90.03 g/mol

Moles H2C2O4 = 0.00428 moles

<u>Step 4: </u>Calculate molarity of H2C2O4

Molarity H2C2O4 = moles / volume

Molarity H2C2O4 = 0.00428 moles / 0.250 L

Molarity H2C2O4 = 0.01712 M

<u>Step 5:</u> Calculate molarity of NaOH

2*Ca*Va = n*Cb*Vb

⇒ with Ca = Molarity of H2C2O4 = 0.01712 M

⇒ Va = volume of H2C2O4 = 0.250 L

⇒Cb = molarity of NaOH = TO BE DETERMINED

⇒ Vb = volume of NaOH = 0.1237 L

Cb = (2*0.01712*0.250)/0.1237

Cb = 0.0691 M

The molarity of the sodium hydroxide solution is 0.0692 M

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(a) It is given that magnesium is reacted with hydrochloric acid and the hydrogen evolved is collected at top. This means that hydrochloric acid will be present in a solution (HCl + Water) and the solvent will be water.

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Total pressure = $P_H_{2} + P_{water}$

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or, $P_H_{2}$ = 746-19.8

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(b) To calculate volume at STP, we will first calculate at $22^{o}C$ and 726.2 mm Hg and than convert it to STP conditions.

Therefore, to calculate volume at $22^{o}C$ and 726.2 mm Hg we will make use of ideal gas law as follows.

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V = 31 ml = $31 \times 10^{-3}$ Litre

According to the ideal gas law ,

PV = nRT

where, P = pressure of the system ,

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R = 0.0821 liter atm/mole K

T = Temperature

Hence, putting the given values into the above formula as follows.

$0.955 \times 31 \times 10^{-3} = N \times 0.0821 \times 295.15$

N = $1.222 \times 10^{-3}$ moles

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Now, at STP T = 273.15 K , P = 1 atm and N = $1.222 \times 10^{-3}$ moles

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= 27.398 ml

Therefore, volume of hydrogen at STP is 27.398 ml .

(c) Now, we can write the the reaction for this case as follows.

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As, weight of magnesium = 0.028 grams

Molar mass of magnesium = 24.3 grams/mole

Number of moles of magnesium = $\frac{mass}{\text{molar mass}}$

= $\frac{0.028}{24.3}$

= $1.15226 \times 10^{-3}$ moles

Since, it can be seen from the reaction that

1 mole of Magnesium = 1 mole of hydrogen

and, moles of hydrogen = $1.15226 \times 10^{-3}$ moles

= 0.001523 moles

Hence, theoretical number of moles of hydrogen that can be produced from 0.028 grams of Mg is 0.001523 moles

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