Question

A student titrated 20 ml of 0.410 m hcl with 0.320 m naoh. determine the volume of naoh needed at equivalence point

Answer 1

Answer:

25.6mL NaOH

Explanation:

We are given the Molarity of the solution ($\frac{moles}{liters}$) and the volume of the solution (.02L).

By multiplying the two together, we can find the moles of solution that are reacted with HCl.

$moles = \frac{.410 moles}{L} *.02L$

This gives us .0082 moles of HCl.

We then find the moles of NaOH that are needed to react with the HCl using the equation.

$HCl + NaOH = NaCl + H_{2} O$

As HCl and NaCl have a 1:1 ratio, we need .0082 mol of NaOH.

Dividing this value by the Molarity of the solution

$\frac{.0082mol}{.320mol/L}$

Gives us the answer, in Liters (.0256), which we can then divide by 100 convert to mL.