Question

A) What is the pH of a solution created by combining 1.28 mole of HOBr and 0.52 mole NaOH in a 1.00L solution? K_{a HOBr} = 2.5 x 10^-9

B) What is the pH of a buffer made by combining 0.62 moles HOBr with 0.88 moles NaOBr in a 100.00mL solution? K_{a HOBr} = 2.5 x 10^-9

Answer 1

Answer – A)

We are given, moles of HOBr = 1.28 moles

Moles of NaOH = 0.52 moles , volume = 1.00 L , Ka = 2.5*10^-9

When we added NaOH in HOBr then there is formed conjugate base of HOBr

HOBr + NaOH -------> OBr^- + H₂O

1.28        0.52                 0.52    

So NaOH is limiting reactant, so moles of OBr^- = 0.52 moles

So moles of HOBr remaining = 1.28-0.52 = 0.76 moles

New molarity –

[HOBr] = 0.76 moles / 1.00 L = 0.76 M

[OBr^-] = 0.52 moles / 1.00 L = 0.52 M

Now we need to calculate pKa of HOBr from given Ka

We know,

pKa = - log Ka

       = - log 2.5*10^-9

      = 8.60

Now we need to use the Henderson Hasselbalch equation-

pH = pKa + log [Conjugate base] / [acid]

pH = 8.60 + log 0.52 M / 0.76 M

     = 8.60+ (-0.165)

      = 8.44

B) We are given, moles of HOBr = 0.62 moles

Moles of NaOBr = 0.88 moles , volume = 100 mL , Ka = 2.5*10^-9

So, molarity of HOBr = 0.62 mole / 0.100 L = 6.2 M

Molarity of [OBr^-] = 0.88 / 0.100 L = 8.8 M

So using the Henderson Hasselbalch equation-

pH = 8.60 + log 8.8 M / 6.2 M

     = 8.60 + 0.1521

      = 8.75