State the relationship of A to B (i.e. <, >, = or can't be determined).

1. Using a 0.05 M Ag+ titrant
A.VEP in titration 25-mL of 0.01 M MgCl2
B.VEP in titrating 25-mL of 0.01 M AlCl3

2. A 25.0-mL solution contains 0.015 M NaCl and 0.010 M HCl
A.VEP (Mohr)
B.VEP (phenolphthalein)

3. pKsp of AgI < pKsp of AgSCN
A.pAg break in the titration of I- with Ag+
B.pAg break in the titration of SCN- with Ag+

4. A 20.0-mL sample contains equimolar amounts of HBr and H2SO4
A.VEP using 0.010 M NaOH
B.VEP using 0.020 M AgNO3

5. EDTA titrant buffered at pH 10
A.[Y4-]
B.[H4Y]

6. KAY > KBY
For a successful EDTA titration
A.Minimum pH of A titration
B.Minimum pH for B titration

7. At the equivalence point for a complexometric titration of Cu2+ buffered at pH 10
A.[Cu2+] with 0.10 M NH3
B.[Cu2+]

8. For the titration of 40-mL of 0.50 M Mn+ with 0.25 M EDTA buffered at pH 10
A.VEP for Mg2+
B.VEP for Al3+

9. For the following electrode: H2C2O4 (aq) + 2H+ (aq) + 2e- ⇌ 2 HCO2H (aq) E° = 0.204 V
A.ECELL at pH 2.0
B.ECELL at pH 3.0

10. A solution containing M2+ is titrated with a standard N+ solution
ERED N2+/N+ > ERED M2+/M+
A.ECELL before EP
B.ECELL after EP

To determine the relationship of A to B for each question, we need to compare the given conditions or parameters for both A and B. Let's go through each question one by one:

1. Using a 0.05 M Ag+ titrant
A. VEP in titration 25-mL of 0.01 M MgCl2
B. VEP in titrating 25-mL of 0.01 M AlCl3

To compare A and B, we need to know the reaction or indicator involved in each titration. The VEP (volume equivalent point) is the volume of titrant required to reach the equivalence point. Without this information, we can't determine the relationship of A to B.

2. A 25.0-mL solution contains 0.015 M NaCl and 0.010 M HCl
A. VEP (Mohr)
B. VEP (phenolphthalein)

Again, without knowing the reactions involved or the color change indicators, we can't determine the relationship of A to B.

3. pKsp of AgI < pKsp of AgSCN
A. pAg break in the titration of I- with Ag+
B. pAg break in the titration of SCN- with Ag+

Based on the given information, we know that pKsp(AgI) < pKsp(AgSCN). The pAg break is related to the solubility product constant (Ksp). Since pKsp(AgI) < pKsp(AgSCN), we can determine that the concentration of Ag+ required to reach the pAg break for A (titration of I- with Ag+) will be higher than that for B (titration of SCN- with Ag+). Therefore, we can state that A > B.

4. A 20.0-mL sample contains equimolar amounts of HBr and H2SO4
A. VEP using 0.010 M NaOH
B. VEP using 0.020 M AgNO3

Without additional information about the reactions involved or indicators used, we can't determine the relationship of A to B.

5. EDTA titrant buffered at pH 10
A. [Y4-]
B. [H4Y]

Without further context or information about the specific EDTA titration and its conditions, we can't determine the relationship of A to B.

6. KAY > KBY
For a successful EDTA titration
A. Minimum pH of A titration
B. Minimum pH for B titration

Based on the given information, we know that KAY (equilibrium constant for the formation of the metal-Y complex) is greater than KBY (equilibrium constant for the formation of the metal-B complex). For a successful EDTA titration, the formation of the metal-Y complex is favored over the formation of the metal-B complex. Therefore, the minimum pH required for the A titration (involving the metal-Y complex) will be higher than the minimum pH required for the B titration (involving the metal-B complex). Therefore, we can state that A > B.

7. At the equivalence point for a complexometric titration of Cu2+ buffered at pH 10
A. [Cu2+] with 0.10 M NH3
B. [Cu2+]

Without additional information about the reactions involved or the specific complexometric titration, we can't determine the relationship of A to B.

8. For the titration of 40-mL of 0.50 M Mn+ with 0.25 M EDTA buffered at pH 10
A. VEP for Mg2+
B. VEP for Al3+

Without knowledge of the reactions and indicators involved, we cannot determine the relationship of A to B.

9. For the following electrode: H2C2O4 (aq) + 2H+ (aq) + 2e- ⇌ 2 HCO2H (aq) E° = 0.204 V
A. ECELL at pH 2.0
B. ECELL at pH 3.0

Given the half-reaction and the standard electrode potential (E°), we can calculate the cell potential (ECELL) at different pH values using the Nernst equation. The formula for the Nernst equation is:

ECELL = E° - (0.0592/n) * log(Q)

where E° is the standard electrode potential, n is the number of electrons transferred, and Q is the reaction quotient. By substituting the given values and calculating ECELL for each pH value, we can compare the cell potentials and determine the relationship of A to B.

10. A solution containing M2+ is titrated with a standard N+ solution
ERED(N2+/N+) > ERED(M2+/M+)
A. ECELL before EP
B. ECELL after EP

Given the reduction potentials (ERED) for the N2+/N+ and M2+/M+ couples, we can compare them to determine the relationship of A to B. If ERED(N2+/N+) is greater than ERED(M2+/M+), we can conclude that the reduction of N+ to N2+ occurs at a higher potential than the reduction of M+ to M2+. Therefore, A (ECELL before EP) would be greater than B (ECELL after EP).