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igor_vitrenko [27]
3 years ago
12

How many milliliters of 5.0 M NaOH are needed to exactly neutralize 40. milliliters of 2.0 M HCl?

Chemistry
1 answer:
zubka84 [21]3 years ago
5 0

Answer:

16mL

Explanation:

Using the following formula;

CaVa = CbVb

Where;

Where

Ca = concentration/molarity of acid (M)

Va = volume of acid (mL)

Cb = concentration/molarity of base (M)

Vb = volume of base (mL)

According to the information provided in this question;

Ca (HCl) = 2M

Cb (NaOH) = 5M

Va (HCl) = 40mL

Vb (NaOH) = ?

Using CaVa = CbVb

Vb = CaVa/Cb

Vb = 2 × 40/5

Vb = 80/5

Vb = 16mL

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Answer:

\boxed {\boxed {\sf 0.3 \ mol \ NaNO_3}}

Explanation:

Molarity is a measure of concentration in moles per liter.

molarity= \frac{moles \ of \ solute}{liters \ of \ solution}

The molarity of the solution is 1.2 M NaNO₃ or 1.2 moles NaNO₃ per liter. There are 0.25 liters of the solution. The moles of solute are unknown, so we can use x.

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We are solving for x, so we must isolate the variable, x. It is being divided by 0.25 liters. The inverse of division is multiplication, so we multiply both sides by 0.25 L.

0.25 \ L *1.2 \ mol \ NaNO_3/L=\frac{x}{0.25 \ L} *0.25 \ L

0.25 \ L *1.2 \ mol \ NaNO_3/L=x

The units of liters cancel, so we are left with the units moles of sodium nitrate.

0.25  *1.2 \ mol \ NaNO_3=x

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There are 0.3 moles of sodium nitrate.

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