Enough of a monoprotic acid is dissolved in water to produce a 0.0136 m solution. the ph of the resulting solution is 2.45. calculate the ka for the acid.

Explanation:

initially conc.         c                              0         0

At eqm.                  

The expression for dissociation constant is,

And,

Now put all the given values in this expression, we get

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

i believe the answer is 5,000