Find the pH of solution (100 mL) of maleic acid XXXXXXXXXXg) treated with 0.259M solution (50 mL) of...

Question

Find the pH of solution (100 mL) of maleic acid XXXXXXXXXXg) treated with 0.259M solution (50 mL) of KOH. Further find concentrations of different forms of maleic acid at the solution. pKa 1 and 2 of maleic acid are 1.92 and 6.27  respectively.

Robert · Accepted Answer

Maleic acid molecular weight = 116.072 
Mass of maleic acid used = 0.753 g 
Number of moles of maleic acid = mass/MW = 0.753 / 116.072 = 0.0064873 moles 
Volume of maleic acid solution: 100 mL  
Molarity of maleic acid= moles/V = 0.0064873/0.1 = 0.064873 M 
Molarity of KOH, M = 0.259 M 
Volume of KOH used, V = 50 mL 
mmoles of KOH used = MV = 0.259 x 50 = 12.95mmoles = 0.01295 moles of KOH used 
At equilibrium: 
H2M + OH- ======== HM- + H2O……………..pKa1 = 1.92 
HM- + OH- ========= M2- + H2O …………….pKa2 = 6.27 
pH = pKa2 + log [M2-]/[ HM-] 
For neutralization for every 1 mole of maleic acid (H2M) two moles of KOH is required.  
For complete neutralization of 0.0064873 moles of maleic acid (H2M),

Find the pH of solution (100 mL) of maleic acid XXXXXXXXXXg) treated with 0.259M solution (50 mL) of KOH. Further find concentrations of different forms of maleic acid at the solution. pKa 1 and 2 of...

Solution

Answer To This Question Is Available To Download

Related Questions & Answers

Submit New Assignment