Chemistry, 31.07.2019 03:30 kathleensumter5100

A25.00 ml sample of an aqueous solution of ba(oh)2 requires 18.45 ml of 0.01500 m hcl (aq) for its neutralization. what is the molarity of the ba(oh)2 solution?

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Answer from: breana758

5.54 x 10⁻³ M

Further explanation

Given:

Neutralization reaction between:

25.00 ml of Ba(OH)₂18.45 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

$\boxed{ \ M = \frac{n}{V} \ }$ $\rightarrow \boxed{ \ n = MV \ }$

$\boxed{ \ Ba(OH)_2 \rightarrow n = (x \ \frac{mol}{L})(25 \ ml) = 25x \ mmol \ }$

$\boxed{ \ HCl \rightarrow n = (0.015 \ \frac{mol}{L})(18.45 \ ml) = 0.277 \ mmol \ }$

Step-2: neutralization

We use the ICE table to see how neutralization occurs between acid and base.

Balanced reaction:

$\boxed{ \ Ba(OH)_2_{(aq)} + 2HCl_{(aq)} \rightarrow BaCl_2_{(aq)} + 2H_2O_{(l)} \ }$

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.

$\boxed{ \ 25x - \frac{1}{2}(0.277) = 0 \ }$

$\boxed{ \ 25x = \frac{1}{2}(0.277) \ }$

$\boxed{ \ 50x = 0.277 \ }$

$\boxed{ \ x = \frac{0.277}{50} \ }$

$\boxed{ \ x = 5.54 \times 10^{-3} \ }$

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⁻ ionsValence of acid = the number of H⁺ ions

Neutralization: $\boxed{ \ V_b \cdot M_b \cdot valence \ of \ base = V_a \cdot M_a \cdot valence\ of \ acid \ }$

$\boxed{ \ (25.00 \ ml) \cdot x \cdot 2 = (18.45 \ ml) \cdot (0.015 \ M) \cdot 1 \ }$

$\boxed{ \ x = \frac{(18.45 \ ml) \cdot (0.015 \ M)}{(25.00 \ ml) \cdot 2} \ }$

Thus the same results were obtained. The molarity of Ba (OH) ₂ solution is 5.54 x 10⁻³ M.

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Answer from: Quest

jjj89t'ji; it hink the answer is your mom is not si

Answer from: Quest

the balanced equation for reaction of ca metal with o2 to produce cao is as below

2 ca+ o2→ 2 cao2

explanation

the reaction is balanced since the number of atoms in reactant side is equal to number of atoms in product side. for example they are 2 atom of ca in reactant side and 2 atoms of ca in the product side.

the process is a meet the requirement of redox reaction because : calcium is oxidized from oxidation state 0 to 2+ while oxygen is reduced from oxidation state 0 to 2-

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A25.00 ml sample of an aqueous solution of ba(oh)2 requires 18.45 ml of 0.01500 m hcl (aq) for its n...

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