Respuesta :
5.54 x 10⁻³ M
Further explanation
Given:
Neutralization reaction between:
- 25.00 ml of Ba(OH)₂
- 18.45 ml of 0.015 M HCl
Question:
What is the molarity of the Ba(OH)₂ solution?
The Process:
Let us say the molarity of the Ba (OH)₂ solution as x M.
Step-1: prepare moles for each reagent
[tex]\boxed{ \ M = \frac{n}{V} \ }[/tex] [tex]\rightarrow \boxed{ \ n = MV \ }[/tex]
[tex]\boxed{ \ Ba(OH)_2 \rightarrow n = (x \ \frac{mol}{L})(25 \ ml) = 25x \ mmol \ }[/tex]
[tex]\boxed{ \ HCl \rightarrow n = (0.015 \ \frac{mol}{L})(18.45 \ ml) = 0.277 \ mmol \ }[/tex]
Step-2: neutralization
We use the ICE table to see how neutralization occurs between acid and base.
Balanced reaction:
[tex]\boxed{ \ Ba(OH)_2_{(aq)} + 2HCl_{(aq)} \rightarrow BaCl_2_{(aq)} + 2H_2O_{(l)} \ }[/tex]
Initial: 25x 0.277 - -
Change: - ¹/₂ · (0.277) -0.277 +¹/₂ · (0.277) +0.277
Equlibrium: - - +¹/₂ · (0.277) +0.277
- Neutralization causes no excess of hydrogen or hydroxide ions in solution. In the end, the number of acid and base reactions is balanced. In other words, the two reagents have run out with nothing left.
- HCl acts as a limiting reagent.
Step-3: calculate the molarity of the Ba(OH)₂ solution.
We consider Ba (OH) from the initial, change, and equilibrium stages.
[tex]\boxed{ \ 25x - \frac{1}{2}(0.277) = 0 \ }[/tex]
[tex]\boxed{ \ 25x = \frac{1}{2}(0.277) \ }[/tex]
[tex]\boxed{ \ 50x = 0.277 \ }[/tex]
[tex]\boxed{ \ x = \frac{0.277}{50} \ }[/tex]
[tex]\boxed{ \ x = 5.54 \times 10^{-3} \ }[/tex]
Thus, the molarity of the Ba(OH)₂ solution is 5.54 x 10⁻³ M.
_ _ _ _ _ _ _ _ _ _
Alternative Steps
- Valence of base = the number of OH⁻ ions
- Valence of acid = the number of H⁺ ions
Neutralization: [tex]\boxed{ \ V_b \cdot M_b \cdot valence \ of \ base = V_a \cdot M_a \cdot valence\ of \ acid \ }[/tex]
[tex]\boxed{ \ (25.00 \ ml) \cdot x \cdot 2 = (18.45 \ ml) \cdot (0.015 \ M) \cdot 1 \ }[/tex]
[tex]\boxed{ \ x = \frac{(18.45 \ ml) \cdot (0.015 \ M)}{(25.00 \ ml) \cdot 2} \ }[/tex]
Thus the same results were obtained. The molarity of Ba (OH) ₂ solution is 5.54 x 10⁻³ M.
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The molarity of the 25 mL Ba(OH)₂ solution required to neutralize 18.45 mL 0.01500 M HCl solution is 0.00554 M
- We'll begin by calculating the number of mole of HCl in the solution. This can be obtained as follow:
Molarity of HCl = 0.015 M
Volume = 18.45 mL = 18.45 / 1000 = 0.01845 L
Mole of HCl =?
Mole = Molarity x Volume
Mole of HCl = 0.015 × 0.01845
Mole of HCl = 0.00027675 mole
- Next, we shall determine the number of mole of Ba(OH)₂ needed to react with 0.00027675 mole of HCl. This can be obtained as follow:
2HCl + Ba(OH)₂ —> BaCl₂ + 2H₂O
From the balanced equation above,
2 moles of HCl reacted with 1 mole of Ba(OH)₂
Therefore,
0.00027675 mole of HCl will react with = [tex]\frac{0.00027675}{2}\\\\[/tex] = 0.000138375 mole of Ba(OH)₂
Thus, 0.000138375 mole of Ba(OH)₂ is needed for the reaction.
- Finally, we shall determine the molarity of Ba(OH)₂ solution. This can be obtained as follow:
Mole of Ba(OH)₂ = 0.000138375 mole
Volume = 25 mL = 25 / 1000 = 0.025 L
Molarity of Ba(OH)₂ =?
Molarity = mole / Volume
= 0.000138375 / 0.025
= 0.00554 M
Therefore, the molarity of the Ba(OH)₂ solution is 0.00554 M
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