Question

Calculate the change in pH when 1.7×10−3 mol of HNO3 is added to 0.100 L of a buffer solution that is 5.2×10−2 M in HF (Ka=3.5⋅10−4) and 9.6×10−2 M in NaF. Express your answers using three significant figures separated by a comma. pH1,pH2 =

Answer 1

1)

Lets calculate the initial pH

Ka = 3.5*10^-4

pKa = - log (Ka)

= - log(3.5*10^-4)

= 3.456

use:

pH = pKa + log {[conjugate base]/[acid]}

= 3.456+ log {9.6*10^-2/5.2*10^-2}

= 3.722

2)

Lets calculate the final pH

mol of HNO3 added = 0.0017 mol

F- will react with H+ to form HF

Before Reaction:

mol of F- = 0.096 M *0.1 L

mol of F- = 0.0096 mol

mol of HF = 0.052 M *0.1 L

mol of HF = 0.0052 mol

after reaction,

mol of F- = mol present initially - mol added

mol of F- = (0.0096 - 0.0017) mol

mol of F- = 0.0079 mol

mol of HF = mol present initially + mol added

mol of HF = (0.0052 + 0.0017) mol

mol of HF = 0.0069 mol

Ka = 3.5*10^-4

pKa = - log (Ka)

= - log(3.5*10^-4)

= 3.456

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

pH = pKa + log {[conjugate base]/[acid]}

= 3.456+ log {7.9*10^-3/6.9*10^-3}

= 3.515

3)

pH change = |final pH - initial pH|

= |3.515 - 3.722|

= 0.2075

Answer:

pH1 = 3.72

pH2 = 3.52