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2 years ago

Determine the number of milliliters of 0.00300 M phosphoric acid required to neutralize 40.00 mL of 0.00150 M calcium hydroxide.

Chemistry

1 answer:

a_sh-v [17]2 years ago

8 0

The volume of H₃PO₄ : 13.33 ml

<h3>Further explanation</h3>

Given

0.003 M Phosphoric acid-H₃PO₄

40 ml of 0.00150 M Calcium hydroxide-Ca(OH)₂

Required

Volume of H₃PO₄

Solution

Acid-base titration formula

Ma. Va. na = Mb. Vb. nb

Ma, Mb = acid base concentration

Va, Vb = acid base volume

na, nb = acid base valence (amount of H⁺/OH⁻)

H₃PO₄⇒3H⁺ + PO₄³⁻ ⇒ 3 H⁺ = valence = 3

Ca(OH)₂⇒Ca²⁺ + 2OH⁻⇒ 2 OH⁻ = valence = 2

Input the value :

a = H₃PO₄, b = Ca(OH)₂

0.003 x Va x 3 = 0.0015 x 40 x 2

Va = 13.33 ml

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Before any reaction occurs, the concentration of A in the reaction below is 0.0510 M. What is the equilibrium constant if the co

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Explanation:

There was a part missing. I think this is the whole question:

<em>Before any reaction occurs, the concentration of A in the reaction below is 0.0510 M. What is the equilibrium constant if the concentration of A at equilibrium is 0.0153 M?</em>

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<em>Remember to use correct significant figures in your answer. Do not include units in your response.</em>

First, we have to make an ICE Chart, which stands for initial, change and equilibrium. We will call "x" unknown concentrations.

A (aq) ⇌ 2B (aq) + C (aq)

I 0.0510 0 0

C -x +2x +x

E 0.0510-x 2x x

Since the concentration at equilibrium of A is 0.0153 M, we get

$0.0510 - x = 0.0153 \\x = 0.0357$

We can use the value of x to calculate the concentrations at equilibrium.

$[A]e = 0.0153 M \\[B]e = 2x = 2(0.0357) = 0.0714 M \\[C]e = x = 0.0357 M \\$

The equilibrium constant, Kc, is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients.

$Kc = \frac{[B]^{2} \times [C]}{[A]} = \frac{0.0714^{2} \times 0.0357}{0.0153} = 0.0119$