Question

1) 7. 10.0 mL of 0.10 M HBr is mixed with 10.0 mL of 0.10 M HCOOH. What is the pH?

2) 15. Calculate the pH of a solution containing 2.5 x 10 –2 mol of nicotinic acid (a monoprotic acid dissolved in 350 mL of water. ( K a = 1.1 x 10 -5 )

3) A 1.0 L buffer solution is made up of 0.15M NaF and 0.20 M HF ( pK a = 3.17) . 0.05 mol of HCl is added to it. What is the new pH? ( Correct answer is 2.77)

Answer 1

VinmL 0 + X 1) In presence of strong and, weak acid ionization will decreases and the concentration of 4* ion from HCOOH is neglisible [487] = verything - Prob 005 PH=-109 (0.05) = 1.30 ?> molarity of nicotinic acid = molesx 1000 (2.5 4162) (400) = 0.0114 M anume nicotinic acid = HA HA + HO HO tao initial 0.0714 +* * change Cooth-x) equilibrium ka = (4,8][AB] [HA] TOOTH-x 2 = 0.00088 M . pH = -log(0.00088) =>3.06 3) HF+ NaF is an acidic buffer, by the addition of Hel, concentration .. of HF increases and concentration of Naf decreases. ndes of HF = 0.20 moles, moles of Naf = 0.15 ndes moles of HF after he addition = 0.20 +0.05 0.25 moes modes of Nafotter Hel addition = 0.15-0.05 0.10 moles pH e pkatlog (Salt) * 3117 +log(0.60) 1.1x165 - 2.77