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4 years ago

Compound a has a pka of 7 and compound b has a pka of 10. compound a is how many times more acidic than compound b?

Chemistry

2 answers:

Pavel [41]4 years ago

4 0

Ka is the acid dissociation equilibrium constant. The larger the value of the Ka, the stronger is the acid. To find Ka from pKa, the equation is:

pKa = -log[Ka]

@pKa = 7
7 = -log[Ka]
Ka = 1×10⁻⁷

@pKa = 10
10 = -log[Ka]
Ka = 1×10⁻¹⁰

This, pKa 7 is more acidic than pKa 10. The scale factor would be:
1×10⁻⁷/1×10⁻¹⁰ = 1,000

<em>Therefore, Compound A is 1,000 times more acidic than Compound B.</em>

oksian1 [2.3K]4 years ago

3 0

Compound A is 1000 times more acidic than compound B

<h3>Further explanation</h3>

The pKa value is an indicator of the strength of an acid

The pKa is a logarithmic function. pKa is the negative logarithm to the base ten of Ka

pKa = - log Ka

The smaller the pKa, the stronger the acid

A difference of 1 pKa shows a difference of 10 times in the acidity

Suppose there is a difference of 3 pKa units, then corresponds to a 10³ difference in acidity

Compound A has a pKa of 7 and compound B has a pKa of 10.

Because the value of pKa compound A is smaller than compound B, compound A is more acidic than compound B

Ka value of each compound:

compound A

pKa = 7

pKa = - log Ka

7 = - log Ka

Ka = 10⁻⁷

compound B

pKa = 10

pKa = - log Ka

10 = - log Ka

Ka = 10⁻¹⁰

So that the difference in acidity

10⁻⁷ / 10¹⁰ = 10³ = 1000 times

<h3>Learn more</h3>

The pH scale is a measurement system

brainly.com/question/599764

the ka reaction of hcn

brainly.com/question/9012584

strong enough acids to react almost completely with a hydroxide ion

brainly.com/question/5273689

Keywords: pKa, strong acid, logarithmic function, compound

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Oksanka [162]

4.48
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25% deprotonated tells us that A- is .25 and that the rest (75% is protonated) thats .75.

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Explanation:

The chemical equation is as follows.

$\frac{1}{2}Cl_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow ClO(g)$

And, the given enthalpy is as follows.

$\frac{1}{2}Cl_{2}(g) + O_{2}(g) \rightarrow ClO_{2}(g)$ ; $\Delta H$ = 102.5 kJ

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Now, total bond enthalpy of per mole of ClO is calculated as follows.

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x = $[(\frac{1}{2})181 + (\frac{1}{2})498] - 243$

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s2008m [1.1K]

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Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:

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<u><em>Using cross multiplication:</em></u>

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Answer:

The answer to your question is:

Explanation:

Quantum numbers have the numbers

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Group of answer choice.

A) 2, -1. 1, -1/2 The option is incorrect "l" could not be negative

B) 2, 1, 0 , 1/2 The option is posible

C) 3, 2, 1, -1 This option is incorrect "s" can not value -1

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I) 0, 2, 0, 1/2 This option is incorrect, n can not value 0

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K) 1, 2, 0, -1/2 This option is incorrect "l" can not value 2.

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