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Calculate the molar concentration of sodium hydroxide solution if 36.00mL of this solution were required to neutralize 25.00 mL of a 0.2000 M solution of hydrochloric acid.
.0288M??????

No. mL x M = mL x M

To calculate the molar concentration of sodium hydroxide solution, we need to use the concept of stoichiometry and the balanced equation of the reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl):

NaOH + HCl -> NaCl + H2O

Given:
- Volume of sodium hydroxide solution (V1) = 36.00 mL
- Volume of hydrochloric acid solution (V2) = 25.00 mL
- Molar concentration of hydrochloric acid (C2) = 0.2000 M

First, we need to determine the number of moles of hydrochloric acid used in the reaction. This can be calculated using the formula:

moles of solute = molar concentration × volume (in liters)

So, moles of HCl (n2) = C2 × V2

n2 = 0.2000 M × (25.00 mL / 1000 mL/ L)
= 0.00500 mol

According to the balanced equation, the mole ratio between hydrochloric acid and sodium hydroxide is 1:1. This means that the number of moles of sodium hydroxide used in the reaction is also 0.00500 mol.

Now, we can calculate the molar concentration of sodium hydroxide (C1). The formula for molar concentration is:

molar concentration = moles of solute / volume (in liters)

So, C1 = n1 / V1

C1 = 0.00500 mol / (36.00 mL / 1000 mL/L)
= 0.1389 M or 0.139 M (rounded to three significant figures)

Therefore, the correct molar concentration of the sodium hydroxide solution is 0.139 M, not 0.0288 M.