Question

find pH. A.) a solution that is 0.170 M in propanoic acid and 0.110 M in potassium propanoate Express your answer using two decimal places. B.) a solution that contains 0.690% C5H5N by mass and 0.990% C5H5NHCl by mass Express your answer using two decimal places. C.) a solution that is 17.0 g of HF and 27.0 g of NaF in 125 mL of solution Express your answer using two decimal places

Answer 1

Answer – A) We are given [CH₃CH₂COOH] = 0.170 M , [CH₃CH₂COONa] = 0.110 M ,

We know Ka for CH₃CH₂COOH = 1.3*10^-5

We need to calculate the pKa from the Ka

we know formula,    

pKa = - log Ka

= - log 1.3*10^-5

         = 4.89

We need to use Henderson Hasselbalch equation-

pH = pKa + log [CH₃CH₂COO^-] / [CH₃CH₂COOH]

pH = 4.89 + log 0.110 / 0.170

     = 4.70

B) We are given, 0.690% C₅H₅N by mass and 0.990% C₅H₅NHCl

We assume, mass of sample = 1.0 g , so mass of C₅H₅N = 0.690 g, mass of C₅H₅NHCl = 0.990 g

So, mass of C₅H₅N = 0.690 g / 79.1 g.mol^-1 = 0.00872 moles

C₅H₅NH⁺ = 0.990 g / 115.56 g.mol^-1 = 0.00857 moles

[C₅H₅N] = 0.00872 moles / 1.0L = 0.00872 M

[C₅H₅NH⁺] = 0.00857 moles / 1.0 L = 0.00857 M

We know pKb for C₅H₅N = 8.77

We know Henderson Hasselbalch equation-

pOH = pKb + log [C₅H₅NH⁺] / [C₅H₅N]

pOH = 8.77 + log 0.00857 / 0.00872

     = 8.76

pH = 14- 8.76

      = 5.24

c) We are given mass of HF = 17.0 g , mass of NaF = 27.0 g , volume = 125 mL

First we need to calculate the moles

HF = 17.0 g / 20.0 g.mol^-1 = 0.85 moles

NaF = 27.0 g / 49.988 g.mol^-1 = 0.643 moles

Molarity of each-

[HF] = 0.85 moles / 0.125 L = 6.8 M

[NaF] = 0.643 moles / 0.125 L = 5.14 M

We know pKa for Hf = 3.14

We know Henderson Hasselbalch equation-

pH = pKa + log [F⁼] / [HF]

pH = 3.14 + log 5.14 / 6.8

     = 3.01