What does the pKa of the conjugate acid of a base have to be to remove a hydrogen from water (pKa = 15.7) so that the ratio of hydroxide (OH-) to water is 100 to 1?

The pKa of the conjugate acid is 17.7

Explanation:

If hydrogen is removed from water, the equilibrium concentration of the conjugate acid according to the information given in the question becomes,

Kₐ = [OH⁻]/[H₂O]

Now, we determine the equivalent pKa

pKa = -log[ka]

pKa = -log[100]

pKa = -2

Removal of hydrogen from water is reversible as shown below;

H₂O  ⇄    OH⁻ + H⁺

15.7             -2

This reaction is reversible, and the difference in pKa = pKa[H₂O] - pKa[H⁺];

pKa  of the conjugate acid = 15.7 - (-2) = 17.7

The pKa of the conjugate acid is 17.7

the electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

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enthalpy change for the reaction between the given volumes of the gases is :