Question

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?

Answer 1

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

Therefore, Compound A is 1,000 times more acidic than Compound B.

Answer 2

Compound A is 1000 times more acidic than compound B

Further explanation

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

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Keywords: pKa, strong acid, logarithmic function, compound