How many milliliters of 4.8 M NaOH are needed to exactly neutralize 29 milliliters of 2 M HCl?

V₁ = 12.1 mL

Explanation:

Given data:

Volume of NaOH required to neutralize = ?

Molarity of NaOH = 4.8 M

Volume of HCl = 20 mL

Molarity of HCl = ?

Solution:

Chemical equation:

HCl + NaOH    →   NaCl  + H₂O

In neutralization reaction the equal amount of  H⁺ from acid and OH⁻ from base react to form water and salt is also formed.

Formula:

M₁V₁ = M₂V₂

M₁ = molarity of NaOH

V₁ = Volume of NaOH

M₂ = molarity of HCl

V₂ = volume of HCl

Now we will put the values.

4.8 M × V₁ = 2 M × 29 mL

4.8 M × V₁ =58 M. mL

V₁ = 58 M. mL / 4.8 M

V₁ = 12.1 mL

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error

c. 4.54 x 10–12 j happens to be correct