Question

Calculate the change in pH to 0.01 pH units caused by adding 10. mL of 3.30-M NaOH is added to 650. mL of each of the following solutions. a) water pH before mixing = pH after mixing= pH change = b) 0.135 M NH41+ pH before mixing = pH after mixing= pH change = c) 0.135 M NH3 pH before mixing = pH after mixing= pH change = d) a buffer solution that is 0.135 M in each NH41+ and NH3 pH before mixing = pH after mixing= pH change =

Answer 1

mmol of NaOH added = MV = 10*3 .3=33 mmol of NaOH:

Q1.

water:

pH before mixing = 7

pH after -->

Vtotal = 10+650 = 660 mL

[NaOH] = mmol of OH- / total V = 33/ 660 = 0.05

pOH = -log(0.05) = 1.30

pH = 14-1.30

pH = 12.7

dpH = (12.7-7) = 5.7

b)

NH4+ is an acid, it has a pKa = 9.25

so

before mixing:

ph = 1/2*(pKa + pC)

pCa = -log(0.135) =0.869

pH = 1/2*(9.25 + 0.869) = 5.06

after mixing:

this is abuffer

pOH = pKb + log(NH4+´/NH3)

pOH = 4.75 + log((0.135*650 - 33)/(33))

pOH = 4.9698

pH = 14-4.9698

pH = 9.03

dpH = 9.03 -5.06 = 3.97

C)

NH3 is a base... when we add NaOH, the OH- is mainly due to NaOH addition

so, simlar to "water"

initial pOH

pOH = 1/2*(pKb + pC)

pOH = 1/2*(4.75+ 0.869) = 2.8095

pH = 14-2.8095 = 11.190

after addition

pOH =

[NaOH] = mmol of OH- / total V = 33/ 660 = 0.05

pOH = -log(0.05) = 1.30

pH = 14-1.30

pH = 12.7

dpH =12.7 - 11.190 = 1.51

d)

for a buffer

pH initial = 9.25 + log(NH3/NH4+)

pH initial = 9.25 + log(0.135/0.135)

pH = 9.25

after 33 mmol o OH:

pH initial = 9.25 + log(NH3/NH4+)

pH initial = 9.25 + log( (650*0.135) + 10 ) / (650*0.135 - 10))

pH mixing = 9.25446

dpH = 9.25446 - 9.25 = 0.0446