Question

1.) What is the pH of a solution created by combining 1.34 mole of HOBr and 0.47 mole NaOH in a 1.00L solution? K_{a HOBr} = 2.5 x 10<sup>-9

2.)</sup> What is the pH after 0.24 moles of NaOH is added to a buffer containing 1.00 moles of NH₃ and 1.00 moles NH₄Cl ? K_{b NH3} = 1.8 x 10<sup>-5

3.)</sup> What is the pH of a buffer made by combining 0.69 moles HOBr with 0.65 moles NaOBr in a 100.00mL solution? K_{a HOBr} = 2.5 x 10^-9

Answer 1

1). Mol HOBr = 1.34 , mol NaOH = 0.47 , volume of solution = 1.00 L

Ka HOBr = 2.5 E-9

We write the reaction between NaOH and HOBr

NaOH (aq) + HOBr(aq)   ------ > NaOBr (aq) + H2O (l)

I           0.47                 1.34                 0                      0

C          -0.47                -0.47                0.47

E          0                      0.87                   0.47  

Here we use Henderson equation

pH = pka + log ([ NaOBr] / [ HOBr] )

pka = -log ka = -log 2.5E-9 = 8.60

pH = 8.60 + log ( 0.47 / 0.87 )

= 8.60

pH = 8.60

We can solve this by using simply the ph = pka ( at half equivalence)

And answer is 8.60

2)

NaOH reacts with NH4Cl as follow

NH4Cl (aq) + NaOH (aq) ---- > NH3 (aq) + H2O(l) + NaCl (aq)

This reaction tells that when x mol of NaOH is added to the buffer,

Final moles of NH4Cl = Initial moles of NH4Cl - x moles of NaOH

Final moles of NH3 = Initial moles of NH3 + x moles of NaOH

Lets use this argument in Henderson equation.

Pkb = -log ( 1.8E-5) =4.74

pOH = pkb + log ([ NH4Cl]/ [NH3])

Lets plug value of final moles of each

pOH = 4.74 + log [( 1.0 – 0.24 ) / ( 1.0 +0.24 ) ]

= 4.53

pH = 14-pOH = 14 - 4.53 = 9.47

pH = 9.47

3). Here we directly use Henderson equation

pH = pkb + log ( [ NaOBr]/[HOBr])

pkb = 8.60

pH = 8.60 + log ( 0.65/0.69)

pH = 8.58

we did not use volume because it cancels from numerator and denominator.