Question
In: Chemistry
Calculate the pH of the titration solution or 25.00 mL of 0.255 M nitrous acid nitrates...
Calculate the pH of the titration solution or 25.00 mL of 0.255 M nitrous acid nitrates with 0.214 M KOH at the following volumes of KOH:
A. 0.00 mL
B. 10.00 mL
C. Mid-point of titration
D. Equivalance point of titration
E. 5.00 mL past equivalence point of titration
Solutions
Expert Solution
Ka of HNO2 = 5.6 x 10^-4
pKa = -log Ka
pKa = 3.25
A. 0.00 mL
pH = 1/2 [pKa -log C]
pH = 1/2 [3.25 -log 0.255]
pH = 1.92
B. 10.00 mL
millimoles of HNO2 = 25 x 0.255 = 6.38
millimoles of KOH = 10 x 0.214 = 2.14
HNO2 + KOH ----------------> KNO2 + H2O
6.38 2.14 0 0
4.235 0 2.14
pH = pKa + log [KNO2 / HNO2]
pH = 3.25 + log (2.14 / 4.235)
pH =2.95
C. Mid-point of titration
here pH = pKa
pH = 3.25
D. Equivalance point of titration
salt only remains
C = 25 x 0.255 / 25 + 29.8
C = 0.116 M
pH = 7 + 1/2 [pKa + log C]
pH = 7 + 1/2 [3.25 + log 0.116]
pH = 8.16
E. 5.00 mL past equivalence point of titration
pH = 12.25
queen_honey_blossom answered 2 years agoRelated Solutions
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