To determine the concentration of 20.00 ml of an unknown solution of a monoprotic acid, it is titrated with 0.1093 m sodium hydroxide. the initial reading on the buret of sodium hydroxide is 19.63 ml. the final reading is 41.63 ml. what is the concentration of the unknown acid? a) 0.09936 mb) 1.000 mc) 0.2275 md) 0.1202 m

d) 0.1202 M

Explanation:

Let's consider the neutralization reaction between NaOH and a generic monoprotic acid.

NaOH + HA → NaA + H₂O

The used volume of NaOH is 41.63 mL - 19.63 mL = 22.00 mL. The moles of NaOH are:

22.00 × 10⁻³ L × 0.1093 mol/L = 2.405 × 10⁻³ mol

The molar ratio of NaOH to HA is 1:1. The moles of HA that reacted are 2.405 × 10⁻³ moles.

The molar concentration of HA is:

2.405 × 10⁻³ mol / 20.00 × 10⁻³ L = 0.1202 M