General Chemistry 4th Edition University Science Books McQuarrie Rock Gallogly p

Question

General Chemistry 4th Edition University Science Books McQuarrie Rock Gallogly presented by Sapling Learning Phosphorous acid, H3PO3(aq), is a diprotic oxyacid that is pKa1 pKa2 1.30 6.70 an important compound in industry and agriculture. lculate the pH for each of the following points in the ration of 50.0 mL of a 3.0 M H3PO3 (aq) with 3.0 M KOH(aq) Number (a) before addition of any KOH Number (b) after addition of 25.0 mL of KOH l Number (c) after addition of 50.0 mL of KOH Number (d) after addition of 75.0 mL of KOH Number (e) after addition of 100.0 mL of KOH HO -PP OH Phosphorous acid Map

Explanation / Answer

a)

before any KOH added, this is mainly a strong acid:

H3PO4 <--> H+ + HPO4- Ka1

since KA1 >> Ka2, we can assume first ionization to bne 100% of ions

so

Ka1 = [H+][HPO4-]/[H3PO4]

10^-1.3 = x*x/(3-x)

x = [H+] = 0.3635

pH = -log(0.3635) = 0.439

b)

mmol of KOH = MV = 25*3 = 75

mmol of H3PO4 = MV = 50*3= 150

this is HALF equivalence point of the FIRST point so

pH = pKa + log(H2PO4- / H3PO4)

pH = 1.3 + log(75/75)

pH = 1.3

c)

after

mmol of KOH = MV = 50*3 = 150

mmol of H3PO4 = MV = 50*3= 150

this is the FIRST equivalance point, this can be clacualted via

pH = 1/2*(pKa1+pKA2)

pH = 1/2*(1.6+6.70) = 4.15

d)

after 75 mL are added... we are in the SECOND half equivalenc epoint

so

pH = pKa2 + log(HPO4-2 / H2PO4-)

so

(HPO4-2 / H2PO4-) = 1

them

pH = pKa2 = 6.70

e)

finally

mmol fo KOH = 100*3 = 300 mmol

mmol of H3pO4 = 50*3 = 150

this is the 2nd euivalence point

this can be calcualted via

pH = 1/2*(pKa2 + pKA3)

pH = 1/2*(6.7+12.33 )

ph = 9.515

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