for the titration of 10.0 ml of .15 M HNO2 aqueous solution (Ka=7.2x10^-4) using

Question

for the titration of 10.0 ml of .15 M HNO2 aqueous solution (Ka=7.2x10^-4) using 2.5 M NaOH aqueous solution, consider the following points of the titration:

A: Initial pH of .15 M HNO2 aqueous solution

B: .1 mL added base

C: 1/2 equivalence point

D: Equivalence point

E Excess base (assume added 100.0 mL of base)

Set up the init mol, final mol, total volume, and ICE table for points B and D only:

Point B:

Titration equation:

init mol:

(assume reaction goes to completion)

final mol:

total volume:

flip reaction equation:

I:

C:

E:

Point D:

Titration equation:

init mol:

(assume reaction goes to completion)

final mol:

total volume:

flip reaction equation:

I:

C:

E:

Explanation / Answer

number of moles of HNO2 = 10 * 0.15 /1000 = 1.5 * 10^-3
pH = 7 + 1/2 (pKa +log C)
pH = 7 + 1/2 (-log(7.2*10^-4) + (2.5*1/(10+1)))
pH = 8.684

pH = pKa at half equivalence point
pH = -log(7.2*10^-4) = 3.143

M1V1 = M2V2
10 * 0.15 = 2.5 * V2
V2 = 0.6 mL
pH = 7 + 1/2(-log(7.2*10^-4)+((2.5*0.6)/(10 + 0.6)))
pH = 8.642

pH = 7 + 1/2(-log(7.2*10^-4) + (((2.5*100)-(0.15*10))/(100-10))))
pH = 9.7

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