Concentration of NaOH 0.1004 M Total amount of dilute phosphoric acide is 10ml T
ID: 988238 • Letter: C
Question
Concentration of NaOH 0.1004 M
Total amount of dilute phosphoric acide is 10ml
Titration 1
Volume of Titrant at First Equivalence Point 5.4ml
What is the moles of H3PO4 Titrated?
Volume of Titrant at Second Equivalence Point 9.5 ml
What is the moles of H3PO4 Titrated?
How do I find out the amount of moles of H3PO4 for both equivalence point?
Determination of pka for the second Ionization of H3PO4
Volume of titrant at the Second Half Equivalence point is 7.5 ml
ph at Second Eq. point is 7.1
What is the pka for the second Ionization of H3PO4?
Please ask me , if you need more information.
CHEMIO17 & CHEM1937 Exp. 4 Titration of a Triprotic Acid Introduction In this experiment you will titrate phosphoric acid (H,PO, a triprotic weak hydroxide (NaOH, a and also the pK, for the second ionization of phosphoric acid. Phosphoric acid can exist in four distinct states of ionization (H,PO4, HPO, HPO and PO, and each of these states is interconverted by the loss or addition of a proton. The three equilibria that describe these saes are defined by the following equations xperiment you will titrate phosphoric acid (H,POs, a triprotic weak acid) with sodium Equation 3 HPOH HP042,H.PO43 Equation 3 Each of these equilibria is described by an acid disassociation consiant Just as groson concentration can be quantified by pH = Aog H ], the acid disassosiauut oossa tombe quantified by pk,Jo, Vealues of pk, ae sseld because thwey c e dinealy adlad values of pH.Explanation / Answer
1 ) Moles of H3PO4 at 1st equi. point = 10.15 / 98 = 0.10 moles [ grams = ml x density]
1 ) Moles of H3PO4 at 2nd equi point = 17.86 / 98 = 0.18 moles i.e gm of H2PO3 = 5.4 x 1.88 = 10.15 ]
pH =pKa
pKa = 7.1
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