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3 years ago

what is the molarity of a solution of sodium hydroxide if it 15.0 mL requires 25.0 ml of 0.25M HCl to reach the end point

Chemistry

1 answer:

DaniilM [7]3 years ago

6 0

Answer:

0.42 M

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

HCl + NaOH —> NaCl + H₂O

From the balanced equation above,

The mole ratio of acid, HCl (nₐ) = 1

The mole ratio of base, NaOH (n₆) = 1

Finally, we shall determine the molarity of the base, NaOH. This can be obtained as follow:

Volume of acid, HCl (Vₐ) = 25 mL

Molarity of acid, HCl (Mₐ) = 0.25 M

Volume of base, NaOH (V₆) = 15 mL

Molarity of base, NaOH (M₆) =?

MₐVₐ / M₆V₆ = nₐ/n₆

0.25 × 25 / M₆ × 15 = 1

6.25 / M₆ × 15 = 1

Cross multiply

6.25 = M₆ × 15

Divide both side by 15

M₆ = 6.25 / 15

M₆ = 0.42 M

Therefore, the molarity of the base, NaOH is 0.42 M

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Answer:

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Explanation:

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Balance charge: $NO_{3}^{-}+4H^{+}+3e^{-}\rightarrow NO+2H_{2}O$ ...............(2)

[ $3\times$ Equation-(1)] + [ $2\times$ Equation-(2)]:

$3ZnS+2NO_{3}^{-}+8H^{+}\rightarrow 3Zn^{2+}+3S+2NO+4H_{2}O$

Oxidation number of S increases from (-2) to (0) for the conversion of ZnS to S. Therefore S is oxidized.

Oxidation number of N decreases from (+5) to (+2) for the conversion of $NO_{3}^{-}$ to NO. Therefore N is reduced.

$NO_{3}^{-}$ consumes electron from ZnS. Therefore $NO_{3}^{-}$ is the oxidizing agent and ZnS is the reducing agent.

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More on stoichiometric problems can be found here: brainly.com/question/14465605

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