A0.146 m solution of a monoprotic acid has a percent dissociation of 1.55%. determine the acid ionization constant (ka) for the acid.

= 3.56 × 10^-5

A monoprotic acid is represented as HA,

Thus;  HA →H+ + A-  

Therefore; Ka = [H+] × [A-] / [HA]  

[H+] = [A-] = Ca ×α  

[HA] = Ca ×(1 - α)  

α = dissociation % / 100 = 0.0155

Thus;

Ka = (0.146 × 0.0155)^2 / (0.146 × (1 - 0.0155))

    = 3.56 × 10^-5

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol