Question

In a titration, a 25.00 mL sample of sodium hydroxide solution was neutralized by 32.72 mL of hydrochloric acid. The molarity of the acid is 0.129 M. Find the concentration of the basic solution.

Answer 1

To find the concentration of the sodium hydroxide solution, we applied stoichiometry to the neutralization reaction with hydrochloric acid. By calculating the moles of HCl and using the 1:1 molar ratio, we determined that the concentration of the sodium hydroxide solution is 0.1689 M.

Step-by-step explanation:

Calculating the Concentration of Sodium Hydroxide Solution

In the given titration question, we are asked to find the concentration of the sodium hydroxide solution. Since the reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl) is a neutralization reaction, we can apply the concept of stoichiometry to solve the problem. First, we must write the balanced chemical equation for the reaction:

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

Now, we utilize the given molarity of HCl (0.129 M) and its volume (32.72 mL) to calculate the moles of HCl, and since the molar ratio of HCl to NaOH is 1:1, the moles of NaOH is the same. Then, using the volume of NaOH solution (25.00 mL), we can find the concentration of NaOH.

The calculation is as follows:

1. Moles of HCl = Molarity of HCl * Volume of HCl in Liters = 0.129 M * 0.03272 L = 0.0042228 moles.
2. Moles of NaOH = Moles of HCl (because of the 1:1 molar ratio) = 0.0042228 moles.
3. Concentration of NaOH = Moles of NaOH / Volume of NaOH in Liters = 0.0042228 moles / 0.02500 L = 0.1689 M.

Therefore, the concentration of the sodium hydroxide solution is 0.1689 M.