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 1 year ago

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