Question

Determine the volume in mL of 0.12 M NaOH(aq) needed to reach the equivalence (stoichiometric) point in the titration of 27 mL of 0.16 M HF(aq). The K_a of HF is 7.4 x 10^-4

Answer 1

volume of NaOH = 36 mL

Explanation

Moles of HF = (molarity of HF) * (volume of HF)

moles of HF = (0.16 M) * (27 mL)

moles of HF = 4.32 mmol

moles of H⁺ = moles of HF

moles of H⁺ = 4.32 mmol

At equivalence point , moles of OH^- = moles of H⁺

therefore, moles of OH^- = 4.32 mmol

All OH^- comes from NaOH

Moles of NaOH = moles of OH^-

moles of NaOH = 4.32 mmol

volume of NaOH = (moles of NaOH) / (molarity of NaOH)

volume of NaOH = (4.32 mmol) / (0.12 M)

volume of NaOH = 36 mL