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2 years ago

15

25.0 mL of an LiOH solution were titrated with 29.15 mLof a 0.205 M H3PO4 solution to reach the equivalence point. What is the m

olarity of the LiOH solution?

1 answer:

borishaifa [10]2 years ago

7 0

Answer:

0.717 M LiOH

Explanation:

(Step 1)

Calculate the moles of H₃PO₄ using the molarity equation.

29.15 mL / 1,000 = 0.02915 L

Molarity = moles / volume (L)

0.205 M = moles / 0.02915 L

0.00598 = moles

(Step 2)

Convert moles H₃PO₄ to moles LiOH using the mole-to-mole ratio from reaction coefficients.

1 H₃PO₄ + 3 LiOH -----> Li₃PO₄ + 3 H₂O
^ ^

0.00598 moles H₃PO₄ 3 moles LiOH
------------------------------------ x -------------------------- = 0.0179 moles LiOH
1 mole H₃PO₄

(Step 3)

Calculate the molarity of LiOH using the molarity equation.

25.0 mL / 1,000 = 0.0250 L

Molarity = moles / volume (L)

Molarity = 0.0179 moles / 0.0250 L

Molarity = 0.717 M

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Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f

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The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

<u>Answer:</u> The initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+2B\rightleftharpoons C$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b$

where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first trial:

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Expression for rate law for second trial:

$1.1\times 10^{-2}=k(0.30)^a(0.100)^b$ ....(2)

Expression for rate law for third trial:

$2.2\times 10^{-2}=k(0.50)^a(0.050)^b$ ....(3)

Dividing 2 by 1, we get:

$\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1$

Dividing 3 by 1, we get:

$\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^1$ ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}$

Calculating the initial rate of formation of C by using equation 4, we get:

$k=1.2M^{-2}s^{-1}$

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

$\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}$

Hence, the initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

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