Answer to Question #330614 in Organic Chemistry for tarandeep singh

Question

Consider a solution initially containing 0.40 mol fluoride anion and 0.30 mol of hydrogen fluoride (HF). What is the pH after addition of 0.20 mol of HCl to this solution? (HF, Ka = 7.2 × 10(–4))? 

Answer 1

F^- + HCl = Cl^- + HF

n(F^-) = n(F^-)_initial - n(F^-)_reacted = 0.40 - 0.20 =0.20 mol

n(HF) = n(HF)_initial + n(HF)_{produced in eaction} = 0.30 + 0.20 = 0.50 mol

pH = pKa + log₁₀$\dfrac{[F^-]}{[HF]}$ = - log10(7.2*10^-4) + log10(0.20/0.50) = 2.74