University Science Books presented by Sapling Learning pKa PKz Phosphorous acid,

Question

University Science Books presented by Sapling Learning pKa PKz Phosphorous acid, HyPO(aq), is a diprotic oxyacid that is an important compound in industry and agriculture. 1.30 6.70 HO Calculate the pH for each of the following points in the titration of 50.0 mL of a 1.8 M HyPO(aq) with 1.8 M KOH(aq). Number (a) before addition of any KOH Number (b) after addition of 25.0 mL of KOH Number (c) after addition of 50.0 mL of KOH I0 Number (d) after addition of 75.0 mL of KOH Number (e) after addition of 100.0 mL of KOH OPrevious Gweup & View Solution Check Answer·Next Ext about us careers prvacy polkey terms of use contact us h

Explanation / Answer

Answer:

pKa1 = 1.30 = -log[Ka]

Ka1 = 0.05

pKa2 = 6.70

Ka2 = 2 x 10^-7

(a) 50 ml of 1.5 M H3PO4

H3PO4 <===> H2PO4^- + H+ let x amount has dissociated then,

Ka1 = x^2/(1.5-x) = 0.05

x^2 + 0.05x - 0.075 = 0

x = 0.25 M

pH = 0.60

(b) 50 ml of 1.5 M H3PO4 with 25 ml of 1.5 M KOH

moles of H3PO4 = 1.5 x 0.05 = 0.075 mols

moles of KOH = 1.5 x 0.025 = 0.0375 mols

remaining H3PO4 = 0.075 - 0.0375 = 0.0375 mols

moles of H2PO4- = 0.0375 mols

Total volume = 0.05 + 0.025 = 0.75 L

molarity of H3PO4 = moles of H2PO4- = 0.0375/0.075 = 0.5 M

log[base]/[acid] = 0

So, pH = pKa = 1.30

(c) 50 ml of 1.5 M H3PO4 with 50 ml of 1.5 M KOH

moles of H3PO4 = 1.5 x 0.05 = 0.075 mols

moles of KOH = 1.5 x 0.025 = 0.075 mols

moles of H2PO4- = 0.075 mols

Total volume = 0.05 + 0.05 = 0.1 L

moles of H2PO4- = 0.075/0.1 = 0.75 M

H2PO4- + H2O <==> H3PO4 + OH-

let x amount has hydrolyzed,

Kb = Kw/Ka2 = 1 x 10^-14/0.05 = 2 x 10^-13

Kb = x^2/0.75 = 2 x 10^-13

x = [OH-] = 3.87 x 10^-7 M

pOH = -log[OH-] = 6.41

pH = 14-pOH = 7.59

(d) 50 ml of 1.5 M H3PO4 with 75 ml of 1.5 M KOH

moles of H3PO4 = 1.5 x 0.050 = 0.075 mols

moles of KOH = 1.5 x 0.075 = 0.1125 mols

so we have 0.0375 mols of HPO4^2-

and, 0.0375 mols of HPO4-

Total volume = 0.05 + 0.075 = 0.125 L

molarity of HPO4^2- = 0.0375/0.125 = 0.3 M

molarity of HPO4- = 0.0375/0.125 = 0.3 M

So log[base]/[acid] = 0

pH = pKa2 = 6.70

(e) with 100 ml of KOH

we would have all of H3PO4 reacted with KOH to form HPO4^2-

moles of HPO4^2- = 0.075 mols

Total volume = 0.150 L

molarity of HPO4^2- = 0.075/0.150 = 0.5 M

HPO4^2- + H2O <==> H2PO4- + OH-

Kb2 = 1 x 10^-14/2 x 10^-7 = 5 x 10^-8

5 x 10^-8 = x^2/0.5

x = [OH-] = 1.58 x 10^-4 M

pOH = -log[OH-] = 3.80

pH = 14 - pOH = 10.20

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.