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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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...