If 35.40 \(\mathrm{mL}\) of 1.000 \(\mathrm{M} \mathrm{HCl}\) is neutralized by 67.30 mL of NaOH, what is the molarity of the NaOH solution?

Short Answer

Expert verified
Therefore, the molarity of the NaOH solution is 0.526 M.

Step by step solution

01

Identify the balanced chemical equation

From their definitions, hydrochloric acid (HCl) and sodium hydroxide (NaOH) react in a 1:1 ratio to form water (H2O) and sodium chloride (NaCl). This gives us the balanced chemical equation: \( \mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{H_2O} + \mathrm{NaCl} \)
02

Compute for moles of HCl

Using the equation for molarity (M) \( M = \frac{{\text{{moles of solute}}}}{{\text{{liters of solution}}}} \), we can find the moles of hydrochloric acid (HCl). Given that we have 1.000 M HCl and a volume of 35.40 mL or 0.0354 L, plug into equation for molarity and solve for the moles of HCl: \( \text{{moles of HCl}} = M \times \text{{volume}} = 1.000 M \times 0.0354 L = 0.0354 moles \)
03

Calculate moles of NaOH

From the balanced chemical equation, we know that 1 mole of HCl reacts with 1 mole of NaOH. So, we have \( 0.0354 \) moles of NaOH.
04

Compute molarity of NaOH

Using the equation for molarity again, we know the moles of NaOH and the volume of NaOH (67.30 mL or 0.0673 L). Plug in the values to the equation: \( M = \frac{{0.0354 \text{{ moles}}}}{{0.0673 \text{{ L}}}} = 0.526 M \)

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