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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