2- a. Find the conditional formation constant for Mg(EDTA)² at pH 9.00. b. Find the concentration of free Mg²+ in 0.050 M Na2 [Mg(EDTA)] at pH 9.00.
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- 11) Will a 0.15 M solution of NH3 be sufficient to keep Zn2+ from precipitating from a solution of 45.0 mL at a concentration of 1 mM before any EDTA is added if the solution is buffered to pH 11? Zn(OH)2 Zn(NH3)+ Zn(NH3)2* Zn(NH3);²* Zn(NH3)2* Ksp-6.0x10-16 (Bi=102.18) (B2=10443) (B3=10%.75) (Be=10%.70)Q: The solubility product of Zn(OH)2 at 25oC is 3.0 x 10-6 M3. Calculate the solubility of Zn(OH)2 in water in gdm-3. [Ar: Zn = 65.5; O = 16; H = 1] The Mr for Zn(OH)2 is 99.5. When I multiply it with Ksp, I didn't get the same answer as the given answer which is 0.904 gdm-3.5 The solubility product constant for Ni₂P₂O7 is 1.7 x 10-13. Calculate Eº for the process 4- Ni₂P₂07 (s) + 4e¯ ⇒ 2Ni(s) + P₂O7
- 12-6. A 100.0 mL solution of 0.050 0 M M"* buffered to pH 9.00 was titrated with 0.050 0 M EDTA. (a) What is the equivalence volume, Ve, in milliliters? (b) Calculate the concentration of M"+ at V = Ve (c) What fraction (ay+-) of free EDTA is in the form Y at pH 9.00?Consider the dissolution of MnS in water (Kep = 3.0 x 10-14). MnS(s) + H20(1) Mn2+(aq) + HS (aq) + OH(aq) How is the solubility of manganese (II) sulfide affected by the addition of aqueous potassium hydroxide to the system? The solubility will be unchanged The amount of KOH added must be known before its effect can be predicted. The solubility will increase The solubility will decrease O O O O14-8. A 10.0-mL solution of 0.050 0 M A£NO3 was titrated with 0.025 0 M NaBr in the cell S.C.E. || titration solution | Ag(s) Find the cell voltage for 0.1 and 30.0 mL of titrant.
- 20. Assume you are using 0.1 M EDTA to titrate a 0.2 M Ca2t solution (25 mL) at pH 10.0. W molarity of free Ca* ions in the solution after 50 mL of the EDTA is added? K=5.0X101º. A. 2.11 X 106 M B. 8.16 X 10 M C. 4.97 X 106 M D. 8.36 X 10 6 MCalculate AG for the following reaction 2 Fe* (ag) + Br2(aq) 2 Fe *(ag)+2 Br (ag) HaRction OP + C HC - 2T 199 Ca e HA + le RD LIC 627 022 SO r -HO, 80 O, + So, PSO, MO Ma 2 010 M, H M O k Au ho, + 2 . 2O 3M 3RO ILO 100 1.3 1.50 Se - Sa N+ 2 N PASO, P so (r0 -014 -0.23 L36 -0.33 133 1.23 -an 121 10, + S HO Za VO 2+ vo HO AuC NO LIO 0.0 -223 0.91 2 +-P -2.71 Na N Ca + 2-a He -2He 077 -250 -209 M,+eMao O-179 kJ O -61.8 kJ O-359 kJ O -1,210 kJ Stop sha Il Proctorio is sharing your screen. O None of these are correct12-E. Suppose that 0.010 0 M Mn2+ is titrated with 0.005 00 M EDTA at pH 7.00. (a) What is the concentration of free Mn2+ at the equivalence point? (b) What is the quotient [H3Y¯]/[H¿Y²¯] in the solution when the titration is just 63.7% of the way to the equivalence point?
- At 25 °C, a 10.0-mL solution of 0.0500 M AGNO3 was titrated with 0.0250 M NaBr in the cell: S.C.E. titration solution | Ag(s). Find the cell voltage for 0.1 mL and 30.0 mL of titrant if the Ksp of AgBr is 5.0 x 10-13. (ESCE = 0.244 V; Ag* + e -> Ag(s) E° = 0.799 V).What should be the pH of the environment for 50% of Ag+ ions to complex in the cyanide (CN¯) anion equivalent to a solution containing 10-3 M Ag+ ions? pKa for HCN = 9,3 ; logß2 for Ag(CN)2 - = 20,0 How can we INCREASE the complexity rate? Please help.The solubility of La(IO3)3 in 42M NaIO3 solution is 5.6 x 10-12 M. Find Ksp for La(IO3)3. 1.1 x 10-11, 7.6 x 10-11, 4.2 x 10-13, 7.4 x 10-35