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Serggg [28]
2 years ago
13

What is the molarity of a NaOH solution when 22.14 mL of 0.105 M oxalic acid is needed to

Chemistry
1 answer:
jeka57 [31]2 years ago
4 0

The molarity of the NaOH solution required for the reaction is 0.186 M

We'll begin by writing the balanced equation for the reaction

H₂C₂O₄ +2NaOH —> Na₂C₂O₄ + 2H₂O

From the balanced equation above,

The mole ratio of the acid, H₂C₂O₄ (nA) = 1

The mole ratio of the base, NaOH (nB) = 2

From the question given above, the following data were obtained:

Volume of the acid, H₂C₂O₄ (Va) = 22.14 mL

Molarity of the acid, H₂C₂O₄ (Ma) = 0.105 M

Volume of the base, NaOH (Vb) = 25 mL

<h3>Molarity of the base, NaOH (Mb) =? </h3>

MaVa / MbVb = nA/nB

(0.105 × 22.14) / (Mb × 25) = 1/2

2.3247 / (Mb × 25) = 1/2

Cross multiply

Mb × 25 = 2.3247 × 2

Mb × 25 = 4.6494

Divide both side by 25

Mb = 4.6494 / 25

<h3>Mb = 0.186 M</h3>

Therefore, the molarity of the base, NaOH solution is 0.186 M

Learn more: brainly.com/question/25739717

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ΔG o for the reaction H2(g) + I2(g) ⇌ 2HI(g) is 2.60 kJ/mol at 25°C. Calculate ΔG o , and predict the direction in which the rea
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The reaction is not spontaneous in the forward direction, but in the reverse direction.

Explanation:

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

H2(g) + I2(g) ⇌ 2HI(g)     ΔG° = 2.60 kJ/mol

Temperature = 25°C = 25+273 = 298 Kelvin

The initial pressures are:

pH2 = 3.10 atm

pI2 = 1.5 atm

pHI 1.75 atm

<u>Step 2</u>: Calculate ΔG

ΔG = ΔG° + RTln Q  

with ΔG° = 2.60 kJ/mol

with R = 8.3145 J/K*mol

with T = 298 Kelvin

Q = the reaction quotient → has the same expression as equilibrium constant → in this case Kp = [p(HI)]²/ [p(H2)] [p(I2)]

with pH2 = 3.10 atm

pI2 = 1.5 atm

pHI 1.75 atm

Q = (3.10²)/(1.5*1.75)

Q = 3.661

ΔG = ΔG° + RTln Q  

ΔG = 2600 J/mol + 8.3145 J/K*mol * 298 K * ln(3.661)  

ΔG =5815.43 J/mol = 5.815 kJ/mol

To be spontaneous, ΔG should be <0.

ΔG >>0 so the reaction is not spontaneous in the forward direction, but in the reverse direction.

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