A few questions I don't really get and need to see the work for

A 50.0 mL sample of 0.55 M benzoic acid, C6H5COOH, a weak monoprotic acid, is titrated with 0.51 M NaOH. Calculate the pH at the equivalence point. Ka of C6H5COOH = 6.5 multiplied by 10-5.

What is the pH of a solution resulting from the addition of 18.0 mL of 0.15 M HCl to 50.0 mL of 0.12 M NH3? Assume the volumes are additive. Kb of NH3 = 1.8 multiplied by 10-5.

What ratio [CH3NH3+]/[CH3NH2] is needed to prepare a buffer solution with a pH of 8.20 from methylamine, CH3NH2, and methylammonium chloride, CH3NH3Cl? Kb of CH3NH2 = 3.7 multiplied by 10-4.

What is the pH of a solution resulting from the addition of 18.0 mL of 0.15 M HCl to 50.0 mL of 0.12 M NH3? Assume the volumes are additive. Kb of NH3 = 1.8 multiplied by 10-5.

What is the molarity of a methylamine solution, CH3NH2(aq), that has the same pH as 0.0773 M NH3(aq)?

A few comments.

1. Five problems in one post-- four too many.
2. No work at all does not engender help and "I don't get it and need to see the work" is not a high incentive. Instead, post one problem and tell us what you don't understand. Surely you understand some things. Telling us what you know and what you don't understand helps us know where to start with the problem. We can help you better if we know how to explain it. Thanks.
3. Some of the rationale for the above comes from the fact that it can take up to 30 minutes each to type in a problem with all of the exponents etc and that can easily be wasted time if you already know that part of our explanation. Again, thanks.

The secret to doing these titration problems is in knowing what you have at certain points in the titration. For benzoic acid titrated to the equivalence point, the acid has been converted to the salt and one has sodium benzoate. Salts, as you know, hydrolyze. If we call benzoic acid HBz and the benzoate anion Bz, then

Bz^- + HOH ==> HB + OH^-
Kb for Bz^- = (Kw/Ka) = (HBz)(OH^-)/(Bz^-).
Substitute Kw, Ka, (Bz = moles/L) and (HBz)=(OH^-)=x. Solve for x, convert to pOH, then pH.

To solve these questions, we will need to use concepts from acid-base chemistry, including equilibrium constants and calculations involving pH.

1. To calculate the pH at the equivalence point of a titration, we need to identify the point where the moles of acid and base are stoichiometrically equivalent. In this case, benzoic acid is a weak acid and reacts with NaOH in a 1:1 ratio.

To find the equivalence point, we can use the equation: moles of benzoic acid = moles of NaOH.

The moles of benzoic acid can be calculated using the formula: moles = molarity × volume (in liters). In this case, the volume is given as 50.0 mL, so we convert it to liters by dividing by 1000.

Next, we can calculate the moles of NaOH using the same formula: moles = molarity × volume (in liters). The molarity of NaOH is given as 0.51 M and the volume is also given as 50.0 mL.

Once we have the moles of NaOH and benzoic acid, we can calculate the concentration of benzoic acid remaining at the equivalence point. Since it reacts in a 1:1 ratio with NaOH, the moles of benzoic acid remaining will be the initial moles minus the moles of NaOH added.

Finally, to find the pH at the equivalence point, we need to determine the concentration of benzoic acid remaining and use the Ka value given. The dissociation of benzoic acid can be represented as: C6H5COOH ⇌ C6H5COO- + H+.

Using the Ka expression, we can set up an equilibrium expression and solve for [H+]. Then, we can calculate the pH using the equation: pH = -log[H+].

2. To calculate the pH of a solution resulting from the addition of HCl to NH3, we need to calculate the concentration of the resulting solution and use the Kb value given for NH3.

First, calculate the moles of HCl added using the formula: moles = molarity × volume (in liters). In this case, the molarity is given as 0.15 M and the volume is 18.0 mL.

Next, calculate the moles of NH3 using the same formula: moles = molarity × volume (in liters). The molarity of NH3 is given as 0.12 M and the volume is 50.0 mL.

Since the volumes are additive, the total volume of the resulting solution is the sum of the initial volumes of HCl and NH3.

Now, use the moles of NH3 and the total volume to calculate the concentration of NH3 in the resulting solution.

Once we have the concentration of NH3, we can use the Kb value given to set up an equilibrium expression for the reaction: NH3 + H2O ⇌ NH4+ + OH-. From this, we can solve for [OH-] and then calculate the pOH using the equation: pOH = -log[OH-].

Finally, use the equation: pH + pOH = 14, to calculate the pH of the resulting solution.

3. To determine the ratio [CH3NH3+]/[CH3NH2] needed to prepare a buffer solution with a pH of 8.20, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the pKa (which is -log of the Ka) and the ratio of the concentrations of the conjugate acid and base.

The Henderson-Hasselbalch equation is: pH = pKa + log([A-]/[HA]).

In this case, CH3NH2 is the base and CH3NH3+ is the conjugate acid.

Start by converting the given pH value to [H+]. Then, use the equation: [H+] = 10^(-pH), to calculate the concentration of H+.

Next, we can use the Kb value given for CH3NH2 to calculate the pKa using the equation: pKa = 14 - pKb.

Now, substitute the values for pH, pKa, and [H+] into the Henderson-Hasselbalch equation and solve for the ratio [A-]/[HA], which represents [CH3NH3+]/[CH3NH2].

4. To find the molarity of a methylamine solution (CH3NH2) that has the same pH as a given NH3 solution, we can use the pKa and pKb relationship to calculate the pOH of the NH3 solution.

First, convert the given NH3 concentration to [OH-] using the Kb value given.

Next, use the equation: pOH = -log[OH-], to find the pOH of the NH3 solution.

Since pH + pOH = 14, we can then calculate the pH of the CH3NH2 solution by subtracting the pOH of the NH3 solution from 14.

Finally, use the pH value obtained and the Kb value given for CH3NH2 to calculate the concentration of CH3NH2 using the Henderson-Hasselbalch equation, similar to the previous question.

Note: It's important to always double-check the calculations and units to ensure accuracy in these types of calculations.