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1. Number of moles of HBr = Molarity x Volume (L)

= 0.15 M x 0.05 L = 0.0075 mol

Number of moles of KOH = 0.25 M x 0.014 L = 0.0035 mol

Number of moles of HBr unreacted = 0.0075 - 0.0035 = 0.004 mol

Total volume of solution = 50 + 14 = 64 mL = 0.064 L

Since HBr is a strong acid, 0.004 mol HBr will dissociate into 0.004 mol H⁺

Hence [H⁺] = Number of moles / Volume (L)

= 0.004 mol / 0.064 L

= 0.0625 M

Hence pH = - log [H⁺]

pH = - log (0.0625)

pH = 1.2

2. Number of moles of NH₃= Molarity x Volume (L)

= 0.2 M x 0.075 L

= 0.015 mol

Number of moles of HNO₃ = 0.5 M x 0.017 L

= 0.0085 mol

HNO₃ reacts with NH₃ to produce NH₄⁺

Hence mol of NH₃ remaining = 0.015 - 0.0085 = 0.0065 mol

Since HNO₃ is the limiting reagent, the number of moles of NH₄⁺ is equal to the number of moles of HNO₃

Mol of NH₄⁺ produced = 0.0085 mol

Given, Kb of NH₃= 1.8 x 10^-5

Since Ka x Kb = Kw = 10^-14

Hence Ka = Kw / Kb = (10^-14) / (1.8 x 10^-5)

Ka of NH₄^{+ ds}= 5.56 x 10^-10

pKa = - log Ka

pKa = 9.25

Using Henderson-Hasselbalch equation we get

pH = pKa + log [NH₃] / [NH₄⁺]

Putting the values we get,

pH = 9.25 + log (0.0065) / (0.0085)

pH = 9.14

3. Number of moles of CH3COOH = Molarity x Volume

= 0.35 M x 0.052 L

= 0.0182 mol

Number of moles of NaOH = 0.40 M x 0.033 L

= 0.0132 mol

CH3COOH reacts with NaOH to prodcue CH3COO^-

Number of moles of CH3COOH remaining = 0.0182 - 0.0132 = 0.005 mol

Since NaOH is the limiting reagent, hence no. of moles of NaOH = No.

Of moles of CH3COO^-

No. of moles of CH3COO^- = 0.0132 mol

Given Ka = 1.8 x 10^-5

pKa = - log Ka

pKa = 4.745

Using Henderson-Hasselbalch equation we get

pH = pKa + log [CH3COO^-] / [CH3COOH]

Putting the values we get,

pH = 4.745 + log (0.0132) / (0.005)

pH = 5.17