PH of mixing Formic Acid with solid NaOH?

HCOOH + NaOH = HCOONa + H2O 100mL of 0.2M HCOOH contains 0.10L x 0.2M = 0.02mole 10mM of NaOH = 0.01 mole When mixed, only 0.01mole of the acid will be nuetralised by the base. Thereby leaving exces of 0.01mole of acid. pH = -log[H+] = -log(0.01) = 2

As Umar said, you began with 0.02 moles of formic acid and added 0.01 mol of NaOH. When you add the NaOH, it reacts quantitatively with the formic acid forming a stoichiometric amount of formate. So, after the addition, you have 0.01 mole of formate and 0.01 mole of formic acid remaining. So, you can use the Henderson-Hasselbalch equation to calculate the pH of the solution: pH = pKa + log [formate]/[formic acid]

As Umar said, you began with 0.02 moles of formic acid and added 0.01 mol of NaOH. When you add the NaOH, it reacts quantitatively with the formic acid forming a stoichiometric amount of formate. So, after the addition, you have 0.01 mole of formate and 0.01 mole of formic acid remaining. So, you can use the Henderson-Hasselbalch equation to calculate the pH of the solution: pH = pKa + log [formate]/[formic acid]