Rank the following titrations in order of increasing pH at the halfway point to
ID: 538787 • Letter: R
Question
Rank the following titrations in order of increasing pH at the halfway point to equivalence (1 = lowest pH and 5 = highest pH). 100.0 mL of 0.100 M HI by 0.100 M NaOH 200.0 mL of 0.100 M HC_2H_3O_2 (K_a = 1.8 times 10^-5) by 0.100 M NaOH 100.0 mL of 0.100 M HF (K_a = 7.2 times 10^-4) by 0.100 M NaOH 100.0 mL of 0.100 M KOH by 0.100 M HCI 100.0 mL of 0.100 M C_2H_5NH_2 (K_b = 5.6 times 10^-4) by 0.100 M HCI Consider the major species present at the halfway point. From this, you should be able to deduce the order of the pH with only minimal calculations Rank the following titrations in order of increasing pH at the equivalence point of the titration (1 = lowest pH and 5 = highest pH). 200.0 mL of 0.100 M HC_2H_3O_2 (K_a = 1.8 times 10^-5) by 0.100 M NaOH 100.0 mL of 0.100 M NH_3 (K_b = 1.8 times 10^-5) by 0.100 M HCI 100.0 mL of 0.100 M C_2H_5NH_2 (Kb = 5.6 times 10^-4) by 0.100 M HCI 100.0 mL of 0.100 M KOH by 0.100 M HCI 100.0 mL of 0.100 M HF (K_a = 7.2 times 10^-4) by 0.100 M NaOH Consider the major species present at the equivalence point. From this, you should be able to deduce the order of the pH with only minimal calculationsExplanation / Answer
Q1.
half way point --> if this is a buffer, the pH = pKa
so
a)
HI and NAOH, must be pretty acidic, since HI is strong acid
pH = 1-2 approx
b)
pH = pKa = -log(1.8*10^-5) = 4.75
c)
pH = pKa = -log(7.2*10^-4) = 3.14266
d)
this is mostly basic, since KOH is left, so pH > 12
e)
this is a wek base, so pKb = pOH
pOH = -log(5.6*10^-4) = 3.25
pH = 14-3.25
pH = 10 .75
therefore
pH trend;:
A > C > B > E > D
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