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You have prepared an aqueous solution of a monoprotic acid (formula mass = 153.483g/mol) by dissolving 26.816g of the acid in sufficient water to make 500.0mL of solution. The pH of the resulting solution is 4.274. What is Ka for this acid?

Respuesta :

pstnonsonjoku

The Ka of the solution is 8.1 * 10^-9.

What is the pH?

The pH of a solution is the concentration of hydrogen ions in the solution.

Number of moles of the acid =  26.816g/153.483g/mol = 0.175 moles

Concentration of the acid =  0.175 moles/0.5 L = 0.35 M

Now we know that the acid reacts with water as follows;

HA(aq)  + H2O(l) ------> H3O^+(aq) + A^-(aq)

[H3O^+] = Antilog (-4.274)

[H3O^+] = 5.32 * 10^-5 M

Since;

[H3O^+]  = [A^-] = x

Ka = x^2/[HA]

Ka = (5.32 * 10^-5)^2/( 0.35)

Ka = 8.1 * 10^-9

Hence, the Ka of the solution is 8.1 * 10^-9.

Learn more about Ka:https://brainly.com/question/16035742

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