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3 years ago

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

Chemistry

1 answer:

Aleksandr-060686 [28]3 years ago

3 0

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.

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Answer:

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Explanation:

In this problem, marble reacted completely with an excess of HCl, which then reacted with NaOH

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CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

Then the excess HCl reacted with NaOH:

HCl + NaOH → NaCl + H₂O

So first let's calculate the amount of excess HCl that reacted with NaOH:

0.9987 M NaOH * 0.02487 L * $\frac{1molHCl}{1molNaOH}$ = 0.02483 mol HCl

Then we calculate the number of unreacted HCl moles in the original 2 L:

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We substract that number from the initial HCl moles in order to find out how many of them reacted with the piece of marble

Initial HCl moles = 2.00 L * 2.52 M = 5.04 mol HCl

HCl moles that reacted with marble = 5.04 - 4.966 = 0.074 mol HCl

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0.074 mol HCl * $\frac{1molCaCO_{3}}{2molHCl}$ * $\frac{100g}{1molCaCO_{3}}$ = 3.7 g marble

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