Calculate the pH of the following buffer solutions:

a) a solution made by dissolving 2.733g KH2PO4 and 4.508g K2HPO4 in water to give 250 ml of solution.

b)a solution made by mixing 35.00 ml 0.24 M C6H5NH2 with 15.00 ml 0.19 M C6H5NH3CL. The base dissociation constant Kb is 4.00 X 10^-10 for C6H5NH2. Assume that the final volume is the sume of the volumes of the two solutions that are mixed.

Use the Henderson-Hasselbalch equation. If you get stuck explain what it is you don't understand.

I don't understand where to start. Do i turn the grams given in part a) to moles first? then i am not sure what to do

how would you do part b?

you have 20 ml of 0.1 M aqueous solution of the weak base (CH3)N (Kb = 7.4 x 10^-5). This solution will be titrated with 0.1 M HCl.

how many ml of acid must be added to reach the equivalence point?

a) 7.432

To calculate the pH of buffer solutions, we need to use the Henderson-Hasselbalch equation, which is given by:

pH = pKa + log([A-]/[HA])

where pH is the acidic or basic nature of the solution, pKa is the negative logarithm of the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.

Let's calculate the pH of each buffer solution step by step:

a) a solution made by dissolving 2.733g KH2PO4 and 4.508g K2HPO4 in water to give 250 ml of solution.

Step 1: Calculate the concentrations of KH2PO4 and K2HPO4.
To calculate the concentrations, we need to convert the mass of each compound to moles and then divide by the total volume.

For KH2PO4:
Molar mass of KH2PO4 = 39.10 g/mol (K) + 1.01 g/mol (H) + 31.00 g/mol (P) + 16.00 g/mol (O) x 4 = 136.09 g/mol
Moles of KH2PO4 = mass / molar mass = 2.733 g / 136.09 g/mol = 0.02006 mol

For K2HPO4:
Molar mass of K2HPO4 = 39.10 g/mol (K) x 2 + 1.01 g/mol (H) + 31.00 g/mol (P) + 16.00 g/mol (O) x 4 = 174.18 g/mol
Moles of K2HPO4 = mass / molar mass = 4.508 g / 174.18 g/mol = 0.02589 mol

Volume of solution = 250 ml = 0.250 L

Concentration of KH2PO4 = moles / volume = 0.02006 mol / 0.250 L = 0.0802 M
Concentration of K2HPO4 = moles / volume = 0.02589 mol / 0.250 L = 0.1036 M

Step 2: Calculate the pKa value.
The pKa value for the phosphate buffer system can be found in reference sources. The pKa value for the H2PO4-/HPO42- system is 7.21.

Step 3: Calculate the pH using the Henderson-Hasselbalch equation.
pH = pKa + log([A-]/[HA])
= 7.21 + log(0.1036 M/0.0802 M)
= 7.21 + log(1.291)
= 7.21 + 0.110
= 7.32

Therefore, the pH of the buffer solution in part a) is 7.32.

b) A solution made by mixing 35.00 mL 0.24 M C6H5NH2 with 15.00 mL 0.19 M C6H5NH3Cl.

Step 1: Calculate the moles of C6H5NH2 and C6H5NH3Cl.
Moles of C6H5NH2 = concentration × volume = 0.24 M × 0.03500 L = 0.00840 mol
Moles of C6H5NH3Cl = concentration × volume = 0.19 M × 0.01500 L = 0.00285 mol

Step 2: Calculate the new total volume of the solution.
Total volume = volume of C6H5NH2 + volume of C6H5NH3Cl
= 0.03500 L + 0.01500 L
= 0.05000 L

Step 3: Calculate the concentrations of C6H5NH2 and C6H5NH3Cl in the mixture.
Concentration of C6H5NH2 = moles / total volume = 0.00840 mol / 0.05000 L = 0.168 M
Concentration of C6H5NH3Cl = moles / total volume = 0.00285 mol / 0.05000 L = 0.057 M

Step 4: Calculate the pKa value.
The pKa value for C6H5NH2 is given as Kb (base dissociation constant) = 4.00 × 10^(-10).
To find pKa, we can use the formula:

pKa + pKb = 14
pKa = 14 - pKb = 14 - (-log10(4.00 × 10^(-10))) = 14 + 9.4 = 23.4

Step 5: Use the Henderson-Hasselbalch equation to calculate the pH.
pH = pKa + log([A-]/[HA])
= 23.4 + log(0.057 M/0.168 M)
= 23.4 + log(0.3393)
= 23.4 - 0.468
= 22.932

Therefore, the pH of the buffer solution in part b) is approximately 22.932.

Yes, convert g to moles. Look up the Ka value for the acid and convert that to pKa, then plug into the HH and calculate pH.