Please help me fill in the boxes for trial 2 Thank you The v

Please help me fill in the boxes for trial 2. Thank you. The volume used is the net volume of 12.8

3. The molar mass and ionization constant of an unknown weak monoprofic acid Concentration of standardized NaOH titrant 0. Mass Concentration of unknown weak monoprotic acid / - mol/L Trial 2 104.47 Triali 104, 10 1.9G 30.93 +s dwps of Indicatur LO. Q0 mL ml Measured pH of the unknown acid solution Mass of unknown acid solution taken for titration Initial buret reading of NaOH titrant Final buret reading of NaOH titrant Net volume of NaOH Millimoles of NaOH to end point of titration Millimoles of unknown acid in sample Molar concentration of unknown acid solution Calculated molar mass of unknown acid 13.0 13.0 2.92 30.53 13.0 mL 25.5 mL 12. 2 1-26 mmol mmol ml mL 1.30 mmol mmol mol/L mol/L g/mol g/mol Calculation of the ionization constant for an unknown acid from the measured pH of the unknown acid samples. (Average the pH and molar concentration values for the samples of unknown acid you titrated: use the average values in the calculation of the ionization constant.) Average molar concentration of unknown for the samples you titrated- - mol/L

Solution

Ans. Trial 2: Given-

            Molarity of standard NaOH = 0.10 M

            Volume of NaOH consumed to reach titration endpoint = 12.8 mL = 0.0128 L

            Concertation of unknown acid = 2.402 g/ L

            Volume of unknown acid solution taken for titration = 104.47 mL = 0.10447 L

# Step 1: Moles of NaOH consumed = Molarity x Volume of solution in liters

                                                            = 0.10 M x 0.0128 L

                                                            = 0.00128 mol

Since the acid is monoprotic, 1 mol acid requires 1 mol NaOH for its neutralization. So, moles of acid in the 104.47 acid solution must be equal to the moles of NaOH consumed to reach titration end point.

So,

            Moles of acid in 104.47 mL unknown acid solution = 0.00128 mL

# Step 2: Mass of unknown acid in 104.47 mL solution =

                                                            Concertation x Volume of acid solution in liters

                                                            = (2.402 g/ L) x 0.10447 L

                                                            = 0.25093694 g

# Step 3: Molar mass of unknown acid =

(Mass of acid / moles of acid) in 104.47 mL sample

= 0.25093694 g / 0.00128 mol

= 196.044 g/ mol