Question

What volume of 2M NaOH is required to completely neutralize 25mL of 1M HCl?

Answer 1

`\boxed {\boxed {\sf 12.5 \ mL \ NaOH}}`

Step-by-step 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.

`HCl _((aq))+NaOH _((aq)) \rightarrow NaCl _((aq)) + H_2O _((l))`

Now let's perform the calculation. We should use the titration equation or:

`M_AV_A= M_BV_B`

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).

`\bullet M_A= 1 \ M \\\bullet V_A= 25 \ mL \\\bullet M_B= 2 \ M`

Substitute the known values into the formula.

`1 \ M * 25 \ mL = 2 \ M * V_B`

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.

`\frac { 1 \ M * 25 \ mL}{2 \ M }=(2 \ M * V_B)/(2 \ M)`

`\frac { 1 \ M * 25 \ mL}{2 \ M }= V_B`

The units of M cancel.

`\frac { 1 * 25 \ mL}{2 }=V_B`

`\frac {25}{2} \ mL= V_B`

`12.5 \ mL= V_B`

12.5 milliliters of 2 M sodium hydroxide are required to neutralize 25 mL of 1 M hydrochloric acid.