Question

1.) 43.5 mL of 1.94 M hydroiodic acid is added to 15.6 mL of barium hydroxide, and the resulting solution is found to be acidic.

24.0 mL of 1.43 M sodium hydroxide is required to reach neutrality.

What is the molarity of the original barium hydroxide solution?

2.) 32.9 mL of 1.49 M perchloric acid is added to 28.9 mL of sodium hydroxide, and the resulting solution is found to be acidic.

27.4 mL of 1.06 M potassium hydroxide is required to reach neutrality.

What is the molarity of the original sodium hydroxide solution?

Answer 1

2HI + Ba(OH)₂ 2 BaI + H₂O

HI + NaOH NaI + H₂O

24 ml 1.43 M NaOH was required to neutralise the excess acid.

Hence , moles of excess acid (Hydroiodic acid ) = moles of NaOH used = 24*1.43/1000 = 0.03432

Original moles of the acid = 43.4*1.94 /1000 = 0.084196

As one mole acid (HI) reacts with two moles of Barium Hydroxide

Then,

moles of Barium Hydroxide =( original moles of acid - moles of excess acid )/2 = (0.084196 - 0.03432) /2 = 0.0249

Therefore, original concentration of Barium Hydroxide ( molarity) = moles/ volume( L) = 0.0249/15.6*10^-3 = 1.59 M.

2.

HClO₄  + NaOH NaOCl + H₂O

HClO + KOH KOCl + H₂O

Moles of Potassium hydroxide required to neutralise excess acid = 27.4*1.06/1000 = 0.029044

Moles of original acid ( HClO₄) = 32.9*1.49/1000 = 0.049021

Hence, moles of Sodium Hydroxide reacted = moles of original acid - moles of excess acid = 0.049021 - 0.029044 = 0.019977

Now, molarity of original Sodium Hydroxide (NaOH ) solution = 0.01977/volume (L) = 0.01977/28.9*10^-3 = 0.691 M