A100.0 ml solution containing 0.864 g of maleic acid (mw=116.072 g/mol) is titrated with 0.276 m koh. calculate the ph of the solution after the addition of 54.0 ml of the koh solution. maleic acid has pka values of 1.92 and 6.27. ph= at this ph, calculate the concentration of each form of maleic acid in the solution at equilibrium. the three forms of maleic acid are abbreviated as h2m, hm−, and m2−, which represent the fully protonated, intermediate, and fully deprotonated forms, respectively. [m2−]= m [hm−]= m [h2m]=

pH = 1.32

Explanation:

                 H₂M + KOH HM⁻ + H₂O + K⁺

This problem involves a weak diprotic acid which we can solve by realizing they amount  to buffer solutions.  In the first  deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:

So first calculate the moles reacted and produced:

n H₂M = 0.864 g/mol x 1 mol/ 116.072 g  =  0.074 mol H₂M

54 mL x  1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH

it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.

moles H₂M left = 0.074 - 0.015 = 0.059

moles HM⁻ produced = 0.015

Using the Henderson - Hasselbach equation to solve for pH:

ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325

Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.

For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.

the atomic number will stay the same no matter how many electrons it has, so it doesn’t decrease nor increase. it’ll only change with a different number of protons