Question

calculate the ph of one solution containing 0.1 M formic acid and 0.1 M sodium formate before and after the addition of 1mL of 5 M Naoh. how much could the Ph change if the NaOh were added to 1 L of pure water

Answer 1

Pka of formic acid (HCOOH)= 3.75

pH= PKa +log [ Sodium formate/ formic acid] = 3.75

NaOH reacts with HCOOH as HCOOH (aq) + NaOH(aq) --> NaCHO2 (aq) and H2O (l)

Moles of HCOOH= 0.1*1= 0.1 moles

Moles of NaOH= 5*1/1000= 0.005 moles

HCOOH is in excess and the excess is = 0.1 - 0.005 = 0.095

Moles of sodium formate = 0.005 + 0.1 = 0.105 moles of HCOOH= 0.095

volume after mixing = 1 + 5 /1 000=1.005

Concentrations : HCOOH= 0.095/1.005 sodium formate= 0.105/1.005

pH= 3.75+log (0.105/0.095)=3.85

When NaOH is added, molarity is , 1*5= 1000*M

M= 5/1000=0.005

The pH remains the same. Only the molarity of NaOH changes.