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3 years ago

5

The acid dissociation constant ka for an unknown acid ha is 4.57 x 10^-3 what is the base dissociation constant kb for th econju

gate base of the acid anion a-

1 answer:

SashulF [63]3 years ago

4 0

Answer:

2.19 x 10^-12.

Explanation:-

The relation between Ka and Kb for an acid and it's conjugate base is

Ka x Kb = Kw where Kw = ionic product of water.

So Kb = 10^-14 / (4.57 x 10 ^ -3)

= 2.19 x 10^-12

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If 42.7 of 0.208 M hydrochloric acid are needed to completely neutralize a solution of calcium hydroxide, how many grams of calc

Montano1993 [528]

Answer:

0.329 g

Explanation:

In the context of this problem, we have a chemical reaction between hydrochloric acid and calcium hydroxide. HCl is the acid here and calcium hydroxide is the base. Hence, we have an acid-base reaction, also known as neutralization reaction.

In a neutralization reaction, water is produced as a product, as well as a salt that we obtain after we exchange the cations: calcium bonds to chloride and hydrogen bonds to hydroxide (the latter is the formation of water). This means we also produce calcium chloride as a product. The overall reaction represents this as:

$Ca(OH)_2(aq)+2 HCl (aq)\rightarrow CaCl_2 (aq)+2 H_2O (l)$

Firslt of all, we wish to find the number of moles of HCl present. Having molarity and volume, this is done by applying the molarity formula. It states that molarity is equal to the rate between moles and volume:

$c_{HCl}=\frac{n_{HCl}}{V_{HCl}}$

Rearranging for moles of HCl, n:

$n_{HCl}=c_{HCl}V_{HCl}$

Based on stoichiometry of the balanced chemical equation, notice that 1 mole of calcium hydroxide reacts with 2 moles of HCl, meaning:

$n_{Ca(OH)_2}=\frac{1}{2} n_{HCl}=\frac{1}{2}c_{HCl}V_{HCl}$

Now that we have the expression for moles, we may also express moles of calcium hydroxide as the ratio between its mass and molar mass:

$n_{Ca(OH)_2}=\frac{m_{Ca(OH)_2}}{M_{Ca(OH)_2}}$

Using the last two equations, we see that:

$\frac{1}{2}c_{HCl}V_{HCl}=\frac{m_{Ca(OH)_2}}{M_{Ca(OH)_2}}\\\therefore m_{Ca(OH)_2}=\frac{1}{2}c_{HCl}V_{HCl}M_{Ca(OH)_2}$

Substitute the given data, as well as the molar mass of calcium hydroxide:

$m_{Ca(OH)_2}=\frac{1}{2}\cdot0.208 M\cdot0.0427 L\cdot74.093 g/mol=0.329 g$

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What element’s valence electrons are located in 5s25p4?

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The element is TELLURIUM. Tellurium has atomic number 52 with the electronic configuration of [Kr] 4d10 5s2 5p4. The element belongs to group 16 elements on the periodic table. It exhibits various oxidation states: +2, +4, +6, and -2. +4 is its most common oxidation state.

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Calculate the concentration of all species in a 0.160 m solution of h2co3.

enot [183]

H₂CO₃ ⇔ HCO₃⁻ + H⁺

I 0.160 0 0

C -x +x +x

E 0.160-x +x +x

Ka1 = [HCO₃⁻][H⁺] / [H₂CO₃]

4.3 x 10⁻⁷ = x² / (0.160-x) (x is neglected in 0.160-x = 0.160)

x² = 6.88 x 10⁻⁸

x = 2.62 x 10⁻⁴

HCO₃⁻ ⇔ CO₃⁻² + H⁺

I 2.62 x 10⁻⁴ 0 2.62 x 10⁻⁴

C -x +x +x

E 2.62 x 10⁻⁴ - x +x 2.62 x 10⁻⁴ + x

Ka2 = [CO₃⁻²][H⁺] / [HCO₃⁻]

5.6 x 10⁻¹¹ = x(2.62 x 10⁻⁴ + x) / (2.62 x 10⁻⁴ - x)

x = 5.6 x 10⁻¹¹

Thus,

[H₂CO₃] = 0.160 - (2.62 x 10⁻⁴) = 0.16 M

[HCO₃⁻] = 2.62 x 10⁻⁴ - ( 5.6 x 10⁻¹¹) = 2.6 x 10⁻⁴ M

[CO₃⁻²] = 5.6 x 10⁻¹¹ M

[H₃O⁺] = 2.62 x 10⁻⁴ + 5.6 x 10⁻¹¹ = 2.6 x 10⁻⁴ M

[OH⁻] = 3.8 x 10⁻¹¹

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