Please please help with #19 part a., b., and c.!! it’s due on monday and i am not terribly good at chemistry. if you could also give an elementary explanation that would be great too!

Please Please Help With #19 Part A., B., And C.!! Its Due On Monday And I Am Not Terribly Good At Chemistry.

Answers

Answer 1

A. 0.2 mole of HCl is needed for the reaction

B. The molarity of NaOH solution is 0.04 M

C. The molarity of HCl is 0.043 M

Molarity is defined as the mole of solute per unit litre of the solution.

Molarity = mole / Volume

A. Determination of the mole of HCl.

We'll begin by calculating the number of mole of NaOH in the solution.

Volume = 0.1 L

Molarity of NaOH = 2 M

Mole of NaOH =?

Mole = Molarity x Volume

Mole of NaOH = 2 × 0.1

Mole of NaOH = 0.2 mole

Finally, we shall determine mole of HCl.

HCl + NaOH —> NaCl + H₂O

From the balanced equation above,

1 mole of NaOH required 1 mole of HCl.

Therefore, 0.2 mole of NaOH will also require 0.2 mole of HCl.

Thus, 0.2 mole of HCl is needed for the reaction.

B. Determination of the molarity of NaOH.

HCl + NaOH —> NaCl + H₂O

From the balanced equation,

Mole ratio of the acid, HCl (nA) = 1

Mole ratio of the base, NaOH (nB) = 1

From the question given above above,

Volume of acid, HCl (Va) = 0.025 L

Molarity of acid, HCl (Ma) = 0.05 M

Volume of base, NaOH (Vb) = 0.034 L

Molarity of base, NaOH (Mb) =?

Mava / mbvb = na/nb

0.05 x 0.025 / Mb x 0.034 = 1

0.00125 / Mb x 0.034 = 1

Cross multiply

Mb x 0.034 = 0.00125

Divide both side by 0.034

Mb = 0.00125 / 0.034

Mb = 0.04 M

Therefore, the molarity of NaOH is 0.04 M

C. Determination of the molarity of the HCl

From the question given above above,

Volume of base, NaOH (Vb) = 0.054 L

Molarity of base, NaOH (Mb) = 0.1 M

Volume of acid, HCl (Va) = 0.125 L

Mole ratio of the acid, HCl (nA) = 1

Mole ratio of the base, NaOH (nB) = 1

Molarity of acid, HCl (Ma) =?  

Mava / mbvb = na/nb

Ma x 0.125 / 0.1 x 0.054 = 1

Ma x 0.125 / 0.0054 = 1

Cross multiply

Ma × 0.125 = 0.0054

Divide both side by 0.125

Ma = 0.0054/0.125

Ma = 0.043 M

Therefore, the molarity of the HCl is 0.043 M

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Explanation:

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Answers

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Part A

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