Use the Henderson–Hasselbalch equation to calculate the pH of each solution:?

"a solution that contains 1.16% C₂H₅NH₂ by mass and 1.24% C₂H₅NH₃Br by mass"

Refer to: https://faculty.ncc.edu/LinkClick.aspx?fileticket=ZxmmjQwsn1w%3D&tabid=1900
Kb for C₂H₅NH₂ = 5.6 × 10⁻⁴
(Different sources may give different values of Kb, and this may lead to slightly different answer.)

Molar mass of C₂H₅NH₂ = (12.0×2 + 1.0×7 + 14.0) g/mol = 45.0 g/mol
Molar mass of C₂H₅NH₃Br = (12.0×2 + 1.0×8 + 14.0 + 79.9) g/mol = 125.9 g/mol

Assume that the mass of the solution is m g.
No. of moles of C₂H₅NH₂ = (1.16%m g) / (45.0 g/mol) = 0.000258m mol
No. of moles of C₂H₅NH₃⁺ = No. of moles of C₂H₅NH₃Br = (1.24% g) / (125.9 g/mol) = 0.0000985m mol
[C₂H₅NH₃⁺]/[ C₂H₅NH₂] = 0.0000985/0.000258

C₂H₅NH₂(aq) + H₂O(l) ⇌ C₂H₅NH₃⁺(aq) …… 5.6 × 10⁻⁴
Henderson-Hasselbalch equation :
pOH = pKb + log([C₂H₅NH₃⁺]/[ C₂H₅NH₂])
pOH = -log(5.6 × 10⁻⁴) + log(0.0000985/0.000258) = 2.83
pH = pKw - pOH = 14.00 - 2.83 = 11.17


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"a solution that is 14.0 g of HC₂H₃O₂ and 10.5 g of NaC₂H₃O₂ in 150.0 mL of solution"

Refer to: http://www4.ncsu.edu/~franzen/public_html/CH201/data/Acid_Base_Table.pdf
Ka for HC₂H₃O₂ = 1.8 × 10⁻⁵
(Different sources may give different values of Kb, and this may lead to slightly different answer.)

Molar mass of HC₂H₃O₂ = (1.0×4 + 12.0×2 + 16.0×2) g/mol = 60.0 g/mol
Molar mass of NaC₂H₃O₂ = (23.0 + 1.0×3 + 12.0×2 + 16.0×2) g/mol = 82.0 g/mol

Molar mass of HC₂H₃O₂ = (14.0 g) / (60.0 g/mol) = 0.233 mol
Molar mass of NaC₂H₃O₂ = (10.6 g) / (82.0 g/mol) = 0.129 mol
[C₂H₃O₂⁻]/[HC₂H₃O₂] = [NaC₂H₃O₂]/[HC₂H₃O₂] = 0.129/0.233

HC₂H₃O₂(aq) ⇌ C₂H₃O₂⁻(aq) + H⁺(aq) …… Ka = 1.8 × 10⁻⁵
Henderson-Hasselbalch equation :
pH = pKa + log([C₂H₃O₂⁻]/[HC₂H₃O₂])
pH = -log(1.8 × 10⁻⁵) + log(0.129/0.233) = 4.49

HH equation is pH = pKa + log [salt]/[acid] OR pOH = pKb + log [salt]/[bse]
To find pH of first buffer, you will need the Kb of CH5NH2 and the molar mass of C2H5NH2 and molar mass of C2H5NH3Br. Find moles of each by using 1.16 g/MW and 1.24 g/MW and find [ ] of each by dividing the moles by 0.1 (100 ml) because that is what % is; it is grams/100 ml. Then plug these concentrations and the pKb into the above formula and solve for pOH, and the for pH.

To find pH of the second buffer, do the same thing. Find moles of HC2H3O2 (14g/molar mass) and moles of NaC2H3O2 (10.5 g/molar mass) and then divide these moles by 0.15 to molar concentrations. Plug these values along with the pKa for HC2H3O2 into the equation and solve for pH.