Answer:
[tex]\boxed {\boxed {\sf 12.5 \ mL \ NaOH}}[/tex]
Explanation:
In a neutralization reaction, an acid and base react to form salt and water. The hydrogen ions (H+) and hydroxide ions (OH-) combine to form water, while the other spectator ions bond to make salt.
For this reaction, the acid is HCl (hydrochloric acid) and NaOH (sodium hydroxide).
The H (from HCl) and OH (from NaOH) combine to make water (HOH or H₂O). The other ions, Cl and Na combine to form NaCl or sodium chloride. Let's write the neutralization reaction.
[tex]HCl _{(aq)}+NaOH _{(aq)} \rightarrow NaCl _{(aq)} + H_2O _{(l)}[/tex]
Now let's perform the calculation. We should use the titration equation or:
[tex]M_AV_A= M_BV_B[/tex]
where M is the molarity of the acid or base and V is the volume.
We know there are 25 milliliters of 1 molar HCl (acid) and an unknown volume of 2 molar NaOH (base).
[tex]\bullet M_A= 1 \ M \\\bullet V_A= 25 \ mL \\\bullet M_B= 2 \ M[/tex]
Substitute the known values into the formula.
[tex]1 \ M * 25 \ mL = 2 \ M * V_B[/tex]
Since we are solving for the volume of the base, we must isolate the variabel. it is being multiplied by 2 M and the inverse of multiplication is division. Divide both sides by 2 M.
[tex]\frac { 1 \ M * 25 \ mL}{2 \ M }=\frac{2 \ M * V_B}{2 \ M}[/tex]
[tex]\frac { 1 \ M * 25 \ mL}{2 \ M }= V_B[/tex]
The units of M cancel.
[tex]\frac { 1 * 25 \ mL}{2 }=V_B[/tex]
[tex]\frac {25}{2} \ mL= V_B[/tex]
[tex]12.5 \ mL= V_B[/tex]
12.5 milliliters of 2 M sodium hydroxide are required to neutralize 25 mL of 1 M hydrochloric acid.
