A solution is prepared by adding 0.298 mol of a weak monoprotic acid (HA) to a...

A solution is prepared by adding 0.298 mol of a weak monoprotic acid (HA) to a sufficient amount of deionized water to give a total solution volume of 400.0 mL. If the Ka of the weak acid is 3.2 x 10^-5, what is the pH of the solution?

Expert Solution

Molarity of weak monoprotic acid (HA) = number of moles of HA/Volume of solution(in L)

=> 0.298/(400/1000) = 0.745 M

Initial 0.745M 0 0

Final (0.745-x)M x x

assuming (0.745-x) is approximately equal to 0.745

queen_honey_blossom answered 2 years ago

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