Question

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?

Answer 1

The correct answer is 2.1

Step-by-step 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