For the titration of 25.0 mL of 0.150 M pyrincline (C_5 H_5 N) with 0.200 M HNO_

Question

For the titration of 25.0 mL of 0.150 M pyrincline (C_5 H_5 N) with 0.200 M HNO_3 Calculate the pH at the 113 of the way point in this titration (this is where you have add 113 of the total amount of strong acid) Calculate the pH after 130 mL of 0.200 M HNO_3 has been added to the pyridine Calculate the pH at the equivalence point of this titration Calculate the pH after a total 240 ml of 0.200 M HNO_3 has been added to the pyridine What volume of 150 M HCl or 1.30 M NaOH would you add to 400 mL of a buffer that is 0.20 M in H_3PO_4 and 0.13 M in H_2PO_4 to change the pH to 233?

Explanation / Answer

A) pOH = pKb + log(salt)/(Base)
pOH = 8.75 + log((0.2*113)/(0.15*25))
pOH = 9.53
pH = 14 - 9.53 = 4.47
B) pOH = pKb + log(salt)/(Base)
pOH = 8.75 + log((0.2*130)/(0.15*25))
pOH = 9.591
pH = 14 - 9.591 = 4.409
C) at equivalence point salt is formed and it is weak base and strong acid salt
pH = 7 - 1/2(pKb + logC)
for C value, M1V1 = M2V2
0.15*25 = 0.2*V2
V2 = 18.75
pH = 7 - 1/2(8.75 + log(18.75*0.2/(18.75+25))
pH = 3.158
D) pOH = pKb + log(salt)/(base)
pOH = 8.75 + log((0.2*24)/(0.15*24))
pOH = 8.875
pH = 14 - 8.875 = 5.125

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