1.) Determine the pH of the solution when 20.00 mL of 0.1563 M aniline hydrochlo

Question

1.) Determine the pH of the solution when 20.00 mL of 0.1563 M aniline hydrochloride (C6H5NH3+Cl-) is titrated with 15.00 mL of 0.1249 M NaOH. (Ka(aniline)=2.40 x 10-5)
2.)Consider the titration of 25.00 mL of 0.4592M HCl which requires 22.74 mL of KOH. Determine the concentration of KOH. 1.) Determine the pH of the solution when 20.00 mL of 0.1563 M aniline hydrochloride (C6H5NH3+Cl-) is titrated with 15.00 mL of 0.1249 M NaOH. (Ka(aniline)=2.40 x 10-5)
2.)Consider the titration of 25.00 mL of 0.4592M HCl which requires 22.74 mL of KOH. Determine the concentration of KOH.
2.)Consider the titration of 25.00 mL of 0.4592M HCl which requires 22.74 mL of KOH. Determine the concentration of KOH.

Explanation / Answer

Answer – 1) We are given, [C6H5NH3+ Cl-] = 0.1563 M , volume = 20.00 mL , [NaOH] = 0.1249 M , volume of NaOH = 15.00 mL

Ka for C6H5NH3+ = 2.40 x10-5

We need to calculate moles of each

Moles of C6H5NH3+ Cl- = 0.1563 M x 0.020 L

                                  = 0.003126 moles

Moles of NaOH = 0.1249 M x 0.015 L

                      = 0.00187 moles

We know reaction –

C6H5NH3+ + NaOH -----> C6H5NH2 + H2O

0.003126    0.00187           0.00187

So, moles of C6H5NH2 formed is 0.00187

Moles of C6H5NH3+ = 0.003126 – 0.00187 = 0.00125

Total volume = 20.00 +15.00 = 35.00 mL

New molarity –

[C6H5NH3+ ] = 0.00125 moles / 0.035 L = 0.0358 M

[C6H5NH2] = 0.00187 moles / 0.035 L = 0.0535 M

We know Henderson–Hasselbalch equation

pH = pKa + log [C6H5NH2] / [C6H5NH3+]

   we are givn Ka for aniline we need to calculate pKa

we know,

pKa = -log Ka

       = - log 2.40 x10-5

       = 4.62

pH = 4.62 + log 0.0535 / 0.00358

        = 4.79

2) We are given, 25.00 mL of 0.4592M HCl and 22.74 mL of KOH

We know neutralization reaction –

HCl + KOH -----> KCl + H2O

We need to calculate moles of HCl first –

Moles of HCl = 0.4592 M x 0.025 L

                      = 0.01148 moles

form the balanced reaction –

1 moles of HCl = 1 mole so KOH

So, moles of KOH = 0.01148 moles

We are given volume of KOH

We know,

Molarity of KOH = 0.01148 moles / 0.02274 L

                       = 0.5048 M

the concentration of KOH is 0.5048

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