Question

What is the pH of the solution after the additional of 0.020 mol of solid NaOH to a 250.0 mL of buffer made of 0.400 M of HF and 0.800 M of KF assuming solution volume does not change (Ka=7.2 x 10−4)?

Answer 1

Henderson equation is

pH = pKa + log ( [ A-]/ [ HA] )

[ A - ] = [ F- ] = 0.800M

[ HA ] = [ HF ] = 0.400M

Ka of HF = 7.2× 10^-4

pKa of HF = 3.14

No of mole of F- = (0.800mol/1000ml)×250ml = 0.2mol

No of mole of HF = (0.400mol/1000ml)×250ml = 0.1mol

NaOH react with HF

NaOH + HF ------> NaF + H2O

0ne mole of NaOH react with 1mole of HF to give one mole of NaF

Therefore,

0.02mole of NaOH react with 0.02mole of HF to give 0.020mole of F-

No of mole of HF = 0.1 - 0.020 = 0.080

No of mole of F- = 0.2 + 0.020 = 0.220

[ HF ] =( 0.080mole/250ml)×1000ml = 0.320M

[ F- ] = ( 0.220mol/250ml)×1000ml = 0.880M

Therefore ,

pH = 3.14 + log ( 0.880/0.320)

= 3.14 + 0.44

= 3.58