A 1000 mL solution containing 09014 g of maleic acid MW 1 1

A 100.0 mL solution containing 0.9014 g of maleic acid (MW = 1 16.072 g/mol) is titrated with 0.2774 M KOH. Calculate the pH of the solution after the addition of 56.00 mL of the KOH solution. Maleic acid has pKa values of 1.92 and 6.27 Number pH= Tools At this pH, calculate forms of maleic acid ion of each form of maleic acid in the solution at equilibrium. The three d H2M, HM-, and M2, which represent the fully protonated, intermediate, x 102 and fully deprotonateo orms, espectively Number [M+]= Number HM Number

Solution

2)

weak acid = 615 ml of 0.250M

number of moles of weak acid = 0.250M x0.615L= 0.15375 moles

NaOH= 500 ml of 0.130M

number of moles o f NaOH= 0.130Mx0.500L= 0.065 mole

       HA   + OH- ------------------ H2A + A-

0.15375     0.065                   0       0

-0.065      -0.065                           0.065

0.08875       0                              0.065

ka= 7.41x10^-5

-log(Ka) = -log(7.41x10^-5)

PKa= 4.13

PH= PKa + log[A-]/[HA]

PH= 4.13 + log(0.065/0.08875)

PH= 3.99

3)

HCl= 28.0ml of 0.280M

number of moles of HCl= 0.280Mx0.0280L= 0.00784 moles

a) NaOH= 38.0ml of 0.280M

number of moles of NaOH= 0.280Mx0.038L= 0.01064 moles

number of moles of Base is greater than the acid.

remaining number of moles of OH-= 0.01064 - 0.00784= 0.0028 moles

number of moles of OH- = 0.0028 moles

total volume= 28.0+38.0= 66.0 ml = 0.066L

Concentration of OH-= number of moles/volume = 0.0028 /0.066 = 0.0424M

[OH-] = 0.0424M

-log[OH-] = -log[0.0424]

POH= 1.373

PH+POH= 14

PH=14-POH

PH= 14-1.373= 12.627

PH= 12.63.

b)

number of moles of HCl = 0.00784 moles

NaOH= 18.0ml of 0.380M

number of moles of NaOH= 0.380Mx0.018L= 0.00684 moles

number of moles of Acid greaterthan the base.

remainign number of moles of acid = 0.00784 - 0.00684 =0.001 moles

number of moles of H+ = 0.001 moles

total volume= 28.0+18.0 = 46.0ml= 0.046L

[H+] = 0.001/0.046 =0.0217 M

[H+] = 0.0217M

-log[H+] = -log[0.0217]

PH=1.66