A solution of hydrochloric acid of unknown concentration was titrated with 0.16 M NaOH. If a 350-mL sample of the HCl solution required exactly 17 mL of the NaOH solution to reach the equivalence point, what was the pH of the HCl solution?

The correct answer is 2.1

Explanation:

The neutralization reaction involved in the titration is between a strong acid (HCl) and a strong base (NaOH), as follows:

HCl + NaOH → NaCl + H₂O

According to this, 1 equivalent of HCl reacts with 1 equivalent of NaOH. The equivalence point is the point at which the amount of NaOH added reacted completely with the amount of HCl.

moles of acid = moles of base

moles HCl = moles NaOH

Ca x Va = Cb x Vb

Ca = (Cb x Vb)/Va = (0.16 M x 17 mL)/(350 mL) = 7.8 x 10⁻³ M

Since HCl is a strong acid, we calculate the pH of the solution directly from the concentration of acid:

[H⁺]= Ca = 7.8 x 10⁻³ M

pH = -log [H⁺] = -log (7.8 x 10⁻³ M) = 2.11

Therefore, the pH of the HCl solution is 2.1