PART 1) Calculate the pH at the stiochiometric point when 75 mL of 0.097 M hydra
ID: 964299 • Letter: P
Question
PART 1) Calculate the pH at the stiochiometric point when 75 mL of 0.097 M hydrazine is titrated with 0.30 M HCl.
PART 2) Calculate the pH at the stiochiometric point when 50 mL of 0.080 M nitric acid is titrated with 0.31 M NaOH.
PART 3)
Which of the coordinates compounds listed can exhibit cis/trans isomerism (C), fac/mer (F) or neither (N). Enter 4 letters in order: CCFN, NNCF. etc.
A) tetraaminedibromocobalt(III) bromide B) dicarbonylbis(trimethylphosphine)platinum(0) C) tris(ethylenediamine)Nickel(II) D) triaminetrichlorochromium(VI) chlorideExplanation / Answer
Part1:
pKb of hidrazine = 5.5 ----> Kb = 10-5.5 = 3.16x10-6
pKa = 14-5.5 = 8.5 --------> Ka = 10-8.5 = 3.16x10-9
The volume to reach the equivalence point is:
Va = 0.097 * 75 / 0.30 = 24.25 mL
Moles Acid = moles Base = 0.097 * 0.075 = 0.00728 moles
[Base] = 0.00728 / 0.09925 = 0.0734 M
At the equivalence point all the hydrazine has been consumed, and the HCl begins to be in excess, so the predominant species here would be the acid and the conjugate acid of the hydrazine, and the reaction taking place would be the following:
r: N2H5+ --------> N2H4 + H+ Ka = 3.16x10-9
i: 0.0734 0 0
e: 0.0734-x x x
3.16x10-9 = x2 / 0.0734-x --> Ka is small so we can assume that x is small too and 0.0734-x can be neglected to:
3.16x10-9(0.0734) = x2
x = [H+] = 1.52x10-5 M
pH = -log(1.52x10-5)
pH = 4.82
Part 2: This is easier cause we have both acid and base as strong, so the dissociation is complete, let's calculate the volume needed to reach the equivalence, and then, the moles, concentration and final pH:
Vb = 0.080 * 50 / 0.31 = 12.9 mL
Total volume = 50+12.9 = 62.9 mL or 0.0629 L
moles = 0.05 * 0.08 = 0.004 moles
[Base] = 0.004 / 0.0629 = 0.0634 M
pOH = -log(0.0634) = 1.196
pH = 14-1.196 = 12.8
Question 3 post it in a new question thread.
Hope this helps
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