Question

fill in the missing words
a solution of hcl with an unknown concentration was titrated with naoh. 25.00 ml of the hcl solution were titrated. 17.65 ml of a 0.5344 m solution of naoh were needed to reach the equivalence point. the molarity of the hcl solution was □ m.

Explanation:

Step1: Recall the titration formula

For a strong acid (HCl) and strong base (NaOH) titration, at equivalence point, the moles of acid equal moles of base. The formula is \( M_1V_1 = M_2V_2 \), where \( M_1 \) and \( V_1 \) are molarity and volume of HCl, \( M_2 \) and \( V_2 \) are molarity and volume of NaOH.

Step2: Identify the values

\( V_1 = 25.00 \, \text{mL} \), \( M_2 = 0.5344 \, \text{M} \), \( V_2 = 17.65 \, \text{mL} \). We need to find \( M_1 \).

Step3: Rearrange the formula to solve for \( M_1 \)

\( M_1=\frac{M_2V_2}{V_1} \)

Step4: Substitute the values

\( M_1 = \frac{0.5344 \, \text{M} \times 17.65 \, \text{mL}}{25.00 \, \text{mL}} \)
First, calculate the numerator: \( 0.5344 \times 17.65 = 0.5344\times17 + 0.5344\times0.65 = 9.0848 + 0.34736 = 9.43216 \)
Then divide by the denominator: \( \frac{9.43216}{25.00} = 0.3773 \) (rounded to four decimal places)

Answer:

\( 0.3773 \)