Question

During a titration, the following data were collected. A 25 mL portion of an unknown monoprotic acid solution was titrated with 1.0 M NaOH(aq). 65 mL of base solution were required to neutralize the sample. How many moles of acid are present in 3.0 L of this unknown solution?

Answer 1

NaOH and the monoprotic acid are in a 1:1 stoichiometric ratio in the neutralization (acid-base) reaction, meaning that 1 mole of base neutralizes 1 mole of acid.

use: \(M_{acid}V_{acid}=M_{base}V_{base}\), to find the molarity of the acid.

then use: \(M_{acid}=\dfrac{n_{acid}}{L_{solution}}\), to find the moles (n) of acid.

Answer 2

its a multiple choice question and I can't seem to be getting the answer. The choices are
-2.6 moles
-0.9 moles
-8.7 moles
-7.8 moles
-5.6 moles

Answer 3

(0.025 L *\(M_{acid}\)=1.0 M NaOH*0.065 L)


\(M_{acid}\)=\(\dfrac{n_{acid}}{L_{solution}}=2.6 M=\dfrac{n_{acid}}{3\;L}\rightarrow3L*2.6M=7.8\;moles\)