Question

A buffer contains 0.50 mol NH₄⁺ and 0.50 mol NH₃ diluted with water to 1.0 L. How many moles of NaOH are required to increase the pH of the buffer to 10.00? (pK_a of NH₄⁺ = 9.25)

Answer 1

A buffer formula:

pOH = pKb + log(conjugate/base)

n = 0.5 NH4+

n = 0.5 NH3

NOTE that the molarity will be equal to the moles since 1 L is used

pKA = 9.25

pKb = 14-pKa = 14-9.5 = 4.75

pOH = pKb + log(conjugate/base)

pOH = 4.75 + log(conjugate/base)

we want pH to be 10, that is pOH = 4

4 = 4.75 + log(conjugate/base)

log(conjugate/base) = -0.75

(conjugate/base) = 0.1778

NH4+ = 0.1778*NH3

WE need to increase the base to increase pH so...

NH4+ = 0.1778*(0.5) = 0.0889 M or mol

1 mol of NaOH ---> 1 mol of NH4+

0.0889 mol of NaOh will be needed to produce 0.0889 mol of NH4+, which will yield a buffer with pH = 10