It takes 5.54 mL of a 2.18 M NaOH solution to neutralize 6.19 mL of an HCl solution. What is the concentration of the HCl solution? Do not include units in your answer. Round to the tenths place or further.

2 M

Explanation:

NaOH + HCl --> NaCl + H2O,

5.54 mL = .00545 L,

If we consider 2.18 M, M [molarity] = mol/L, so we have 2.18 mol/L. From this we can get the moles of NaOH.

.00545 L *  2.18 mol/L ≈ .0121 mol

The concentration of HCl can be found from the following:

.0121 mol = .00619(x) [where x = concentration of HCl]

x = 1.955 M ≈ 2 M

We equate the two as we require the same amount in order to neutralize the base.

lol for the pointsss

it takes on an electronical energy