IIL. Acid Base Titration (20 pts Answer the following questions for the titratio
ID: 696829 • Letter: I
Question
IIL. Acid Base Titration (20 pts Answer the following questions for the titration of 50.0 mL of 0.10 M of weak monoprotic acid, HA, with 0.05 M of NaOH solution. Ka of HA is 2.0x 10. What is the volume of the NaOH solution required to reach the a. equivalence point? b. Calculate the pH after adding 25.0 mL of the NaOH solution. c. Calculate the pH after adding 125.0 mL of the NaOH solution IV. Complexometric Reactions and Titrations Answer the following question for the titration of Fe with EDTA Write the ionic equation for the reaction of Fe2 with EDTA (H,Y) and write the expression of the formation constant for the reaction (6 pts) 1. Calculate the pFe in 50.0 mL solution of 0.0150 M Fe at pH-7.0 after adding 25.00 mL of 0.0300 M EDTA. Kr of Fe-EDTA chelate is 2.1 x 10 and au at pH 7.0 is 4.8 x 10. (22 pts) 2. V. Gravimetric Analvsis and Presipitation Equilibria The H2S in a 50.0 g sample of crude petroleum was removed by distillation and collected in a solution of CdCl2. The precipitated CdS was then filtered, washed and ignited to Cdsos. Calculate the mass percentage of H S (gram formula weight 34.08 g/mol) in the sample if 0.108 g of CdsO, (gram formula weight 208.48 g/mol) was recovered. (11 pts) 1· 2. a Calculate the molar solubility of Ba(IO) in pure water. Ksp of Ba(IO,)2 is 1.57 x 10 ? (9 pts) solutions of Ba(IO3)h? (4 pts) Ba(NO,). (10 pts) b. What is the molar concentrations of Ba2 and 10, ions in a saturated c. Calculate the molar solubility of Ba(I0,)h in a solution that is 0.010M inExplanation / Answer
III)
a)
find the volume of NaOH used to reach equivalence point
M(HA)*V(HA) =M(NaOH)*V(NaOH)
0.1 M *50.0 mL = 0.05M *V(NaOH)
V(NaOH) = 100 mL
Answer: 100. mL
b)
we have:
Molarity of HA = 0.1 M
Volume of HA = 50 mL
Molarity of NaOH = 0.05 M
Volume of NaOH = 25 mL
mol of HA = Molarity of HA * Volume of HA
mol of HA = 0.1 M * 50 mL = 5 mmol
mol of NaOH = Molarity of NaOH * Volume of NaOH
mol of NaOH = 0.05 M * 25 mL = 1.25 mmol
We have:
mol of HA = 5 mmol
mol of NaOH = 1.25 mmol
1.25 mmol of both will react
excess HA remaining = 3.75 mmol
Volume of Solution = 50 + 25 = 75 mL
[HA] = 3.75 mmol/75 mL = 0.05M
[A-] = 1.25/75 = 0.0167M
They form acidic buffer
acid is HA
conjugate base is A-
Ka = 2*10^-3
pKa = - log (Ka)
= - log(2*10^-3)
= 2.699
we have below equation to be used:
This is Henderson–Hasselbalch equation
pH = pKa + log {[conjugate base]/[acid]}
= 2.699+ log {1.667*10^-2/5*10^-2}
= 2.22
Answer: 2.22
c)
we have:
Molarity of HA = 0.1 M
Volume of HA = 50 mL
Molarity of NaOH = 0.05 M
Volume of NaOH = 125 mL
mol of HA = Molarity of HA * Volume of HA
mol of HA = 0.1 M * 50 mL = 5 mmol
mol of NaOH = Molarity of NaOH * Volume of NaOH
mol of NaOH = 0.05 M * 125 mL = 6.25 mmol
We have:
mol of HA = 5 mmol
mol of NaOH = 6.25 mmol
5 mmol of both will react
excess NaOH remaining = 1.25 mmol
Volume of Solution = 50 + 125 = 175 mL
[OH-] = 1.25 mmol/175 mL = 0.0071 M
we have below equation to be used:
pOH = -log [OH-]
= -log (7.143*10^-3)
= 2.1461
we have below equation to be used:
PH = 14 - pOH
= 14 - 2.1461
= 11.85
Answer: 11.85
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