Question

The student titrated 10 ml of standered 0.15 M HCl with his sodium hydroxide solution. When the titration reached the equivalence point, the student found that he had used
10.3 ml of Sodium hydroxide solution. Calculate the molarity of the sodium hydroxide solution.

Answer 1

Answer:

0.15 M

Explanation:

Step 1: Write the neutralization reaction

NaOH + HCl ⇒ NaCl + H₂O

Step 2: Calculate the moles of HCl that reacted

10 mL of 0.15 M HCl was used. The moles of HCl that reacted are:

$0.010L \times \frac{0.15mol}{L} = 1.5 \times 10^{-3} mol$

Step 3: Calculate the moles of NaOH that reacted

The molar ratio of NaOH to HCl is 1:1. Then, the moles of NaOH that reacted are 1.5 × 10⁻³ moles.

Step 4: Calculate the concentration of NaOH

1.5 × 10⁻³ moles of NaOH are in 10.3 mL of solution. The molarity of NaOH is:

$\frac{1.5 \times 10^{-3} mol}{10.3\times 10^{-3}L} =0.15 M$