plz help fill the blanks 8.5 ml 9.9 ml Volume of Titrant at Second Equivalence P
ID: 988141 • Letter: P
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plz help fill the blanks
8.5 ml 9.9 ml Volume of Titrant at Second Equivalence Point (mL and L Total Moles of NaoH Added at Second Eq. Point (moles) Moles of H3PO4 Titrated at Second Eq. Point (moles) ation of H3PO Solution based on second eq. points of each titration) Average Concentration of H3PO4 Solution MD (from first and second eq. points of both titrations) Determination of the pK, for the Second Ionization of H3PO4 Volume of Titrant at Second Half Equivalence Point (mLO 7.7 m 6.55 ml 70 pH at Second Half Eq. Point 6.8 pKa for Second Ionization of H3PO4Explanation / Answer
Assuming 50 ml of 1 M of H3PO4 is titrated with 1 M NaOH
moles = molarity x volume
molarity = moles/volume
H3PO4 + 1OH- ---> H2PO4- + H2O ------ Ist equivalence point
H3PO4 + 1.5OH- ------> 0.75H2PO4- + 0.75HPO4^2- + 1.5 H2O------- IInd half equivalence point
H3PO4 + 2OH- ----> HPO4^2- + 2H2O ------ IInf equivalence point
moles of NaOH added = moles of H3PO4 ---- at First equivalence point
moles of NaOH added = 1/2 moles of H3PO4 ----- second equivalence point
Volume of NaOH needed to reach second half equivallence point = 3 times volume needed to reach first half Eq point
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Volume of titrant at second Eq. point 9.9 ml (0.0099 L) 8.5 ml (0.0085 L)
Total moles of NaOH added at IInd Eq point (moles) 0.0099 0.0085
Moles of H3PO4 added at IIn Eq pt.(moles) 0.00495 0.00425
conc. of H3PO4 (based on IInd Eq.pt. (M) 0.099 0.085
Avg. conc. of H3PO4 (M) 0.092
pH at second half Eq. point = pKa2 = 7.20
pKa for second ionization of H3PO4 = 7.20
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