Step-by-step explanation:
The statement explains that a titration was carried out in which the titrated (solution whose concentration is unknown) is HCl (acid). The titrant (solution placed in the burette to react with HCl) was NaOH (base). With this titration, the following data were obtained:
MNaOH = ??
VNaOH = 13.33 mL
VHCl = 31.73 mL
MHCl = 0.044 M
To solve this exercise and find the concentration in mol/L of NaOH, the first step is to know how to write the equation that represents the neutralization reaction that occurred and balance it correctly. Neutralization reactions follow the following scheme: acid + base → salt + water.
In this case, the following reaction occurred: NaOH + HCl → NaCl + H2O
Note that the equation is already balanced. Balancing the equation correctly is important because we need to look at the stoichiometric ratio between the reactants, which in this case is 1:1.
The second step is to relate the obtained data. Since a total neutralization reaction took place, this means that the number of H+ ions and OH- ions present in the medium became equal. So we have:
nHCl = nNaOH
Knowing that n = M . V, we can perform the following substitution:
nHCl = nNaOH
MHCl*VHCl = MNaOH*VNaOH
MNaOH = ??
VNaOH = 13.33 mL
VHCl = 31.73 mL
MHCl = 0.044 M
0.044*31.73 = 13.33*MNaOH
MNaOH = 0.1 M
Answer: MNaOH = 0.1 M